—€ 2052200 TOEO O MNO IOHM/18lN Rate iin y H. O. No. 217 Maneuvering Board Manual UNITED STATES NAVY DEPARTMENT HYDROGRAPHIC OFFICE UNITED STATES GOVERNMENT PRINTING OFFICE WASHINGTON : 1941 For sale by the Hydrographic Office, Washington, D.C. - - ~- Price $1.00 Also by the Superintendent of Documents, Washington, D. O. STATUTES OF AUTHORIZATION There shall be a Hydrographic Office attached to the Bureau of Naviga- tion in the Navy Department for the improvement of the means for navigat- ing safely the vessels of the Navy and of the mercantile marine by providing, under the authority of the Secretary of the Navy, accurate and cheap nautical charts, sailing directions, navigators, and manuals of instructions for the use of all vessels of the United States, and for the benefit and use of navigators generally (R. S. 431). The Secretary of the Navy is authorized to cause to be prepared, at the Hydrographic Office attached to the Bureau of Navigation in the Navy Department, maps, charts, and nautical books relating to and required in navigation, and to publish and furnish them to navigators at the cost of print- ing and paper, and to purchase the plates and copyrights of such existing maps, charts, navigators, sailing directions, and instructions as he may consider necessary, and when he may deem it expedient to do so, and under such regula- tions and instructions as he may prescribe (R. 8. 432). It PREFACE The Maneuvering Board Manual is compiled primarily for the purpose of furnishing officers of the U. S. Navy with a printed publication, in an un- restricted status, to which ready reference can be made for aid in the solution of tactical problems. Explanations have, therefore, been simplified and reduced to a minimum consistent with clarity. From time to time, in the past, many different typewritten pamphlets on this subject have been compiled by various Naval officers and much material has also appeared in scattered articles in the Naval Institute Proceedings, but no regular printed publication on this subject has ever been issued. This text is therefore arranged with the assistance of officers attached to the Post- graduate School, United States Naval Academy, Annapolis, Md. The material has been freely drawn from all of the above-mentioned sources, but principally from the problems as given in the Maneuvering Board Manual prepared for students and graduates of the Postgraduate School. Constructive criticism is therefore freely invited of any part of the text or its drawings. G. S. Bryan, Captain, U. S. Navy, Hydrographer. III INTRODUCTION The ability of certain outstanding navigators and tacticians to rapidly and efficiently carry out missions, conduct scouting and search operations, and shift stations within a fleet or other mobile unit has long been known, Although their skill has been described by such terms as ‘‘having developed a good seaman’s eye,” basically their aptitude has been the result of being able to apply the principles of relative movement to the particular problem at hand. Relative movement is an everyday phenome- non. The most familiar example of this is the apparent movement of celestial bodies across the sky. As the globe turns from the West to the Kast, to an observer stationed on the earth, the celestial bodies appear to rise in the East and set in the West. When two trains on adjacent tracks are moving in the same direction but at different speeds, to passengers on the faster train it appears that the slower train is moving backwards. By movement relative to the faster train and ignoring the actual direction and distance traveled over the face of the earth by both, that is what the slower train is doing. The essential difference between the relative movement method of solving problems and the usual navigational plot method, is one of origins. The latter uses a point fixed with respect to the earth and called a “‘Chart Point.” The travel of units, portrayed by lines on the chart used, represents directions and distances actually traversed on the face of the earth or over the ground. Such a diagram, when used in this publication, will be referred to as the “Navigational Plot.” The lines representing the travel of units over the ground in this diagram are called “Chart Lines.” When several units are being plotted on this diagram, their exact positions for any particular time must be carefully delineated before their positions relative to each other can be found. For a composite picture of the actions of several units, this is excellent; for planning actions in advance, the amount of trial and error involved usually causes much delay, so the relative movement method is to be pre- ferred in most cases. The relative movement method uses a moving unit instead of a chart point as the point of origin. The unit so chosen, designated as the “‘Guide,’’ remains fixed in the Relative Plot, although it represents a ship moving over the earth. The 0°00" 10.0 Knots FIGURE 1. FIGURE 2. movement of all other units concerned in the problem is referred to this guide and their travel is portrayed by “Relative Move- ment Lines.” The relative movement line of a maneuvering unit, with respect to the guide, is defined by drawing a line through plotted successive positions of the maneuvering unit, as determined by range finder and compass aboard the guide. These positions are laid off from a fixed point representing the position of the guide in the Relative Plot. The Maneuvering Board has been designed to facilitate this plotting as well as to show graphically all the maneuvers involved. In the relative movement method, two diagrams are normally required, the ‘‘ Vector Diagram’ and the “Relative Plot.” The “Navigational Plot’? may be added as part of the solution or as a check on results. The Vector Diagram is so called because every line in it is a vector and therefore indicates both magnitude and direction. The magnitude of the vector is its length, which applied to the scale in use indicates a velocity, or rate. The direction of the vector is shown by its inclination, an arrow being added to the head of the vector to prevent reciprocal errors. Vector quantities are added and subtracted geometrically by the Parallelogram Law as outlined in standard textbooks. The point of origin in a vector diagram is usually a point fixed with respect to the earth, and for convenience is lettered “e.” ‘Thus, vectors originating at “e’’ show direction and rate of travel with respect to the earth, or over the ground. In Figure 1, three such vectors are shown. e... . g indicates the travel of a unit G, making 10.0 knots in direction 000° over the ground; e . . . . w represents the travel of the wind in direction towards 020° at 12 knots over the ground; while e . . . . m shows the unit M on course 070° at a speed of 9.0 knots over the ground. The Wind Vector indicates the direction toward which the wind is blowing, instead of the usual method of describing the direction from which it originates. These vectors, or any others based on the same units of motion, may be combined to form the Vector Diagram. Figure 2 shows the combination of the above vectors to indicate concurrent travel. By joiningg ... . m another vector is formed. If we consider m as the head and g as the foot of the resultant vector, g . . . . m gives the direction and rate of travel of unit M relative to unit G. In the corresponding Relative Plot, G would be stationary while M traversed a line parallel tog... . mand at the rate indicated by the length of g.... m. If, on the contrary, we consider g as the head and m the foot of this vector, then m . . . . g gives the direction and rate of travel of unit @ relative to unit M, and in the corre- sponding Relative Plot, 14 would remain stationary while G traversed a line parallel to and in direction m . . . . g, at the rate IV indicated by the length of m ... .g. Similarly, joining w and m, and considering m the head and w the foot of the vector so formed, w .... m represents the travel of unit M in respect to the wind. If this vector is reversed and w made the head, then m . . . . w represents the travel of the wind in respect to unit M or shows the apparent wind on M. From the above it will be noted that all lines in the Vector Diagram are vectors, which have in them only the quantities of direction and rate of travel. No other quantities can be directly obtained from this diagram. The Relative Plot is the second diagram employed in the Relative Movement Method. This diagram has a fixed point, representing the unit used as a guide, and such Relative Lines as the solution may produce. These Relative Lines show direction and distance travelled, both relative to the Guide Unit. The Relative Plot, therefore, has only the quantities of direction and distance. The common factor in both the Vector Diagram and the Relative Plot is direction. Hence, the interchange of data from one diagram to the other must be based upon this common quantity. The element of Time, required in practically all problems is the factor by which speed, from the Vector Diagram, is converted to Distance in the Relative Plot or vice versa. The solution for Time alone may be reached by dividing the distance, from the Relative Plot, by the speed, from the Vector Diagram. The two are interchanged through their common quantity of direction. Although the majority of the problems dealt with herein are solved by the Relative Movement Method, recourse is fully had to other methods which more readily yield the desired results. Particular mention must be made of the Navigational Plot, which can always be utilized as a proof of the solution, regardless of what method was actually used in working the problem. A solution by this method may be extremely laborious, involving trial and error, but once the results are obtained by some easier and quicker method, it is a matter of but a few moments to obtain a check by completing the Navigational Plot. In actual practice, whenever time permits, it should be employed not only to check the results, but also to insure that there is no navigational hazard in the direction of your own ship’s motion. It must be pointed out that the mere working out of a problem on the Maneuvering Board is not enough. The progress of your own vessel along the line of relative movement must be checked by both bearings and rangefinder ranges. In some instances the guide may change both course and speed considerably without previous signal. It is far easier to set up another problem based on your instantaneous position than to try to involve the previous solution with the problem presented by the new conditions. All officers should be familiar with the simplest types of problems and their rapid solution. The operator at sea, however, should never become so intent on figuring out his next proper move on a Maneuvering Board that he fails to keep a sharp lookout for incidental ships, such as other maneuvering units, merchant ships, or aircraft carriers, about to launch or land aircraft, which may happen to wander into his set-up. Also, no shoals or menaces to navigation are shown on the Maneuvering Board. A lack of alertness in this regard will never be compensated by the ability to find perfect solutions on a Maneuvering Board. SUGGESTIONS The Maneuvering Board, HO 2665, is especially designed to facilitate the solution of problems by Vector Diagrams and Relative Movement Plots, although any chart equipped with a compass rose may be used if dividers, parallel rulers, and a proper scale are available. When using the Maneuvering Board, place the Vector origin, e, at the center of the diagram. Letter the head and foot of each vector as it is drawn and indicate the head by an arrowhead. Leave the foot plain or enclosed in a small circle. Use small letters in the Vector Diagram to correspond with capital letters designating the same unit in the Relative Plot. Any letters desired by the operator may be used to indicate the various units concerned. In this publication, G is used to designate the Guide unit and g indicates the head of the Guide’s vector. If a single maneuvering unit is involved, it is usually designated M, with the corresponding m as its vector head, but this lettering may be varied for the sake of clarity. In the Vector Diagram, it is suggested that the letters e, g, and w be used as herein. In the Relative Plot, indicating the unit Guide by G will tend to prevent errors. In the figures illustrating the various examples in this publication, distinctive lines are used to emphasize the construction of the set-up. For ordinary solution of problems, this is not necessary. Courses and bearings should be drawn in their true directions. Relative bearings are easily found by orienting the dia- gram. The application of variation and deviation to true courses and bearings after the solution is reached readily yields the magnetic or compass courses and bearings, with less chance for error than if the conversion were attempted earlier in the solution. No set rule can be given for the scale to be used in the Relative Plot. It should be as large as the size of the board will permit. G may be placed at any point, but usually it is more convenient to place it at the center of the board, superimposed upon e. Care should be taken that the Relative Plot and the Vector Diagram are not confused thereby. Sometimes, by plac- ing G elsewhere, a much larger scale can be employed, with consequent increase in accuracy. Care should be exercised to differentiate between the scale used in the Vector Diagram and the scale used in the Relative Plot. These diagrams are entirely separate. The habit of noting in the upper right-hand corner of the board the proper scales used, such as 1 division=2 miles, or 1 division=3 knots, will generally prevent confusion in this matter. Also, when using HO 2665, if the scales on the side are employed, a D over the distance scale used and an S over the speed scale employed, will further tend to reduce error. Confusing the distance scale and the speed scale is the most frequent source of mistake made by the beginner. Finally, the scales chosen should be as large as practicable. Strive for accuracy but not by the expenditure of excessive time. It is much more advantageous for the officer conning to have an approximate course and speed to start with, modified as necessary later for the exact course and speed required, than to be slow in changing station. The standard of accuracy required should be sufficiently exact as to produce results which Vv would be acceptable at sea. Use the correct values for your own course and speed and the best estimated values for the guide’s course and speed, which is not often exactly known. When in formations, cultivate the habit of having the position of all units plotted on the Maneuvering Board, with the formation guide at the center. This removes doubt as to the location of those units which may interfere with the Sapcaitions changing of your own station as well as permits a more rapid solution of the problem presented. The value of this habit especially when on the scouting line, cannot be overestimated. A rough sketch on plain paper may assist in choosing the proper scale for plotting the formation. The use of accurate parallel rulers and a sharp pencil will aid in neatness and clarity. If, however, no parallel rulers are available, two drafting triangles will suffice. When determining the point of tangency between a line and a circle, estimating by eye is insufficient. Determine the direc- tion of the tangent, add or subtract 90° and lay off this direction from the center of the circle. Its intersection with the original line will determine the point of tangency. Follow the latest practice of using three-figure numbers for courses or bearings. If a decimal part of an hour or mile is to be indicated, place a zero before the decimal point. All courses or bearings are true unless otherwise indicated. Finally, remember that while the maneuvering unit travels along the Relative Movement Line, its heading coincides with this line only in the special case in which the course of the maneuvering unit is the same as the guide’s course. In other words the maneuvering unit generally travels along the Relative Movement Line “‘crab fashion.” VI GLOSSARY The following definitions of terms used in this publication are listed for the information of students: Air Course.—The corrected compass course steered by aircraft through the air. By combining the Wind Vector and the known Air Speed, the equivalent Ground Course and Speed are obtained. Air Speed.—The speed of an aircraft relative to the air. This speed, when known, is represented by a circle of proper radius to the scale in use with its center at win the Vector Diagram. This circle is treated in the same manner as ground speed circles drawn with e as their center and used for surface craft. Equivalent Ground Speed is obtained by combining the Air Speed with the Wind Vector. Chart Course.—The true direction over the surface of the earth that an aircraft or ship is intended to travel. Chart Point—A properly located point which is fixed in respect to the ground or earth. Chart Track.—A line representing the true course and distance made good or desired to be made good between stations on the earth. Current Vector.—A vector with origin at e, representing travel of the water over the ground or current. The head of this vector is usually lettered c. It is seldom necessary to use this vector except when anchoring. Constant Bearing Line.—The relative line traversed by a unit whose bearing from the reference unit does not change. This condition exists when the guide unit and the maneuvering unit arrive at a common point, diverge from a common point, or when the two have the same vector. In the first case, this is called the “Collision Bearing.” Guide.—The chosen unit to which movement of other units concerned is referred. In the Relative Plot, the Guide remains stationary. Maneuvering Unit.—Any moving unit set up in the problem except the Guide. Specifically, a Maneuvering Unit is one whose movements are under investigation. Navigational Plot.—A diagram such as is used in ordinary chart or navigational work. Sometimes the Navigational Plot is used as the alternate solution which may be more rapid than the Relative Movement Plot. Its most common usage is to check or prove the results found in the latter. Relative Movement Line.—A line passing through successive relative positions occupied by the Maneuvering Unit plotted from bearings and range-finder distances taken on the Guide, whose position remains stationary in the Relative Plot. It is commonly called a ‘Relative Line.” Relative Plot.—One of the fundamental diagrams used in Relative Movement methods. A fixed point indicates the posi- tion of the Guide and one or more Relative Movement Lines are properly related to this point for the movements of the Maneu- vering Unit. This diagram yields position, as well as direction and distance of relative travel. Relatwe Position —The position occupied by one unit with respect to another, defined by range and bearing of one from the other. The bearing may be either true or relative. Slope.—A direction or inclination. This term is usually used in connection with the transfer of a line parallel to itself from one diagram to another. Speed Cirele.—A circle drawn about a point or the end of a vector with the radius equal to a given speed. Such circles drawn with ¢ as center indicate speed over the ground. If drawn with was center they indicate speed of aircraft through the air. Time Line.—A line joining the heads of two vectors which represent successive courses and speeds of a specific unit in passing from an initial to a final position in known time, via a specified intermediate point. This line also touches the head of a constructive unit which proceeds directly from the initial to the final position in the same time. By general usage this con- structive unit is called the Fictitious Ship, and the head of its vector, marked “‘f,” divides the Time Line into segments inversely proportional to the times spent by the Reference Unit on the first and second legs. The Time Line is used in two-course problems. Vector.—A straight line which indicates by its inclination and length respectively the direction and rate of travel of a unit represented by the head of the vector relative to a unit represented by the foot of the vector. Vector Diagram.—A coordinated set of vectors representing concurrent travel of the units concerned. It is a fundamental diagram in the Relative Movement Method. This diagram yields rate and direction of travel. 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In this section are typical cases based upon variations of this premise. For this reason, the set-up of the problem and the method of solution should be carefully studied. In some instances a choice between two or more Maneuvering Unit Vectors is possible. This arises from the statement of the problem being sufficiently broad, either by accident or design, to permit more than one solution. In actual practice, other factors may be present which will preclude an alternate solution. In the first few examples quotation marks are included for the sake of emphasizing steps. These are omitted in the later examples. Case I TO PROCEED FROM ONE RELATIVE POSITION TO ANOTHER IN A SPECIFIED TIME GIVEN: COURSE AND SPEED OF GUIDE, INITIAL AND FINAL RELATIVE POSITIONS, AND TIME ALLOWED. TO DETERMINE: COURSE AND SPEED OF MANEUVERING UNIT. Example.—Guide ‘‘G”’ is on course 205° true, speed 10.0 knots. The Maneuvering Unit “4”, now 12.0 miles due west of “‘G’’, is ordered to take station 20.0 miles dead ahead of ‘‘G”’, arriving on station in 1.5 hours. Required.—(a) Course for “‘M”’. (b) Speed of ““M”’. (See fig. 3). Procedure.—Plot Guide at ‘‘G”’ and locate initial and final positions of “‘]/7’”’ at M, and M, respectively. JoinM,.... Mb). Draw vector “fe. . . . §g’’, representing course and speed of Guide. Since the Maneuvering Unit moves along the Relative Movement Line M, . . . . Mh, transfer the slope of this line to “6”, From “‘g’’, in the same direction that M, lies from M,, measure a vector equal to the rate of Relative Travel, which is equal to the Relative Distance M, . . . . M, divided by the time available, 1.5 hours. This is found most readily by using the Logarithmic Scale at the bottom of the Maneuvering Board, placing a straight edge along the Relative Distance (18.5 miles) and the time allotted (1.5 hours or 90 minutes). This Relative Speed is found to be 12.4 knots, which is the length of “gs... . m’. Vector “e.. . . m’’ represents the required course and speed of the Maneuvering Unit. Answer.—(a) 185°. (b) 21.4 knots. NOTE.—Had the Maneuvering Unit in this problem been a plane, there would have been no change in the Relative Plot. In the Vector Diagram, however, the Wind Vector, ‘“‘e. . . . w’’ would be laid off from “‘e”’ if it measured the true wind; if it were apparent wind on “‘ G”’ it would be laid offas‘‘é... . w’ from“ sg”. The Air Course and Air Speed would be found by joining ‘“‘ m”’ and ‘‘w”’ instead of ‘‘m’’ and “e” as shown. The true course to be steered would be found by transferring the Slope ‘‘w .. . . m’’ to the compass rose on the diagram. This is illustrated more fully in some of the following examples. Successive positions of the Maneuvering Unit are shown in the Relative Plot of figure 3 in order to stress the important fact that although the maneuvering Unit is travelling down the Relative Movement Line, its heading does not necessarily coincide with the slope of this line. It is, how- ever,-inclined in the direction shown by the vector‘‘e. . . . m’”’. ae Bo SCALES | Division=2 Miles , ee aor | Division=3 Knots *- | 76-95 72-90 63 65 64-80 60-75 56 --70 52-65 48-60 44 55 40-50 36 F-45 32-40 28 35 24 F-30 20-25 16 20 2b 5 8 Ew 4E5 o Eo gs 8 TIME in minutes g 2 g 8 8 8 $ 8 pistaNce in yards Relative or actual Given any two corresponding quantities, solve = i OCS & &RSSS3 £8 __ DISTANCE in miles for third by laying rule through points on prop- er scales and read intersection on third scale. = « o 8 ae ata 2 &§ 8 8 8 8 8 SPEED in knots MANEUVERING BOARD z Sse Sa Weshington, D.C., published May, 1920, at Hydrographic Office, Price) 6Dlcentsi(per pad ot'$0) under the authority of the SECRETARY OF THE NAVY New Publication: 4th. Ed, Apr, 1938 Ath. Edition, April 1938 No. 2685 a FIGURE 3. Case Il TO PROCEED FROM ONE RELATIVE POSITION TO ANOTHER ON A SPECIFIED COURSE GIVEN: COURSE AND SPEED OF GUIDE, INITIAL AND FINAL RELATIVE POSITIONS, AND THE COURSE TO BE USED BY THE MANEUVERING UNIT. TO DETERMINE: SPEED OF MANEUVERING UNIT AND TIME TO ARRIVE IN FINAL POSITION. Example.—Guide on course 230°, speed 10.0 knots. Ship ‘‘M’’, now 4.0 miles on the starboard beam of the Guide is ordered to take station 10.0 miles broad on the port quarter of the Guide, steering course 180° en route. Required.—(a) Speed of “‘M’’. (b) Time for “‘”’ to reach new position. (See fig. 4.) Procedure.—Plot Guide at ‘‘G”’ and locate initial and final positions of Maneuvering Unit at M, and Mb, respectively. Join M, Cu MONS M2. Draw vector “‘e ... . g’’, representing course and speed of ‘‘G’’. From “‘e”’ lay out course line specified for ‘‘”’. Transfer slope ““M, .. . . M,’’ to “‘g’’, intersecting the course line of ‘‘M’’ at ‘“‘m’’. The vector “‘e . . . . m’’ repre- sents the course and speed of the Maneuvering Unit. Divide the Relative Distance. ‘‘M, .... My’, by the Relative Speed, ““g .... m’’, to obtain the time required to reach the new station on the specified course. This is easily done on the Logarithmic Scale. Answer.—(a) 8.82 knots. (b) 98 minutes or 1 hour 38 minutes. NOTE.—Should the Maneuvering Unit be a plane, the wind vector, ‘“‘e .... w’’, would appear in the Vector Diagram. The specified ground course is laid off from ‘‘e”, as above. By reference to‘‘w”’ the Air Speed and Air Course corresponding could be found. If the Air Course were specified, this direction would be laid off from ‘‘ w”’ instead of from “‘e”’, and the proper Air Speed determined by the intersection of this course line with the transferred slope “MM, .... M,’’ laid off from‘‘s’’. Time in either event would be obtained in the manner illustrated. I Divisionst'Milé - "+. | Division=! Knot - + fete fee ag De ee ee oe . Bs) eee eee LOGARITHMIC SCALE n 2 os 3 38 238 8 ssgss 8 8 a 2 2 R 8 8 8 gee38 £8882 8 8 8 88288 8 888: 8 2S 82828 2 8 pistance inyards ; 4 Relative or actual Given any two corresponding quantities, solve = me ON te 2£ & &RSVV3 Rg 8 DISTANCE in miles for third by laying rule through points on prop- er scales and read intersection on third scal2. - ” 2 ow © Nn Oa 8s 2 & 28 & 8 e 2 8 MANEUVERING BOARD i at Relative or actual Price 60 cents (per pad of 50) Washical=ny DiC pobshad Nay) 72), et ipsceraghic, OCH) under the authority of te SECRETARY OF THE NAVY New Publication: 4th. Ed, Apr, 1938 4th. Edition, April 1933 No. 2655 a FIGURE 4. Case Ill TO PROCEED FROM ONE RELATIVE POSITION TO ANOTHER AT A SPECIFIED SPEED GIVEN: COURSE AND SPEED OF GUIDE, INITIAL AND FINAL RELATIVE POSITIONS, AND SPEED TO BE USED BY MANEUVERING UNIT. TO DETERMINE: COURSE OF MANEUVERING UNIT AND TIME TO ARRIVE IN FINAL POSITION. Exam ple.—Guide on course 037°, speed 16.0 knots. Ship ‘‘M”’’, now bearing 060° (true) and distant 11.5 miles from the Guide, is ordered to take position 16.0 miles 30° abaft the starboard beam of the Guide, using a speed of 12.0 knots enroute. Required.—(a) Course or courses for “‘M’’. (b) Time required to reach final position. (See fig. 5.). Procedure.—Plot Guide at “‘G’’, and locate initial and final positions of ‘‘M’’ at M, and M, respectively. M2 bears 120° relative from “G” or 157° (true) from ‘“‘G”’. Join M,.... ™M. Draw vector “‘e .... g’’, representing course and speed of “‘G’’. With “‘e” as center and with radius equal to the specified speed for ‘‘M’’ (12.0 knots), describe a circle. This circle, of course, will contain all the courses available to ‘‘M”’ at the specified speed, no matter what the other requirements. Transfer the slope WM, .... M, to “‘g’’, cutting the speed circle of “‘M”’ at points ‘‘m,”’ and ‘‘m,’’. “e.... mm,’ and “e.... m,’’arevectorsthat “‘M”’ could use. An inspection of the diagram quickly reveals the fact that the latter vector yields the greatest Relative Speed; therefore ‘‘M” would naturally choose the course indicated by this vector unless there were other vessels or obstacles present which would dictate the use of the vector “‘e . . . . my,’’. The time required to arrive at the final position is found by dividing the Relative Distance by the Relative Speed as usual. For vector “fe . . . . m,’’ and using the Logarithmic Scale this is shown graphically. Answer.—(a) 153° or alternate course 047°. (b) 52 minutes by most expeditious course or 4.43 hours by steering course 047°. NOTE.— Only one solution can be obtained if the speed of the Maneuvering Unit is greater than the speed of the Guide. Two solutions are other- wise possible unless the specified speed for the Maneuvering Unit is the minimum speed that would allow him to reach the final position. In case the Maneuvering Unit is a plane, draw the wind vector “‘e” .... w’’, and about “‘w”’ draw the speed circle with radius equal to the 6699 given air speed of the plane. This speed circle is treated in precisely the same manner as the ground speed circle, drawn about “e” in figure 5. Since the air course would be desired in this case, it is obtained by reference to “‘w”’ instead of ‘‘e”’. . 5 SCALES _ I Division=2'Milés , 4:15:21 iasl Division =2 Knots *-_ 76 95 72-90 63 Fas 64 80 60-75 56 70 52 F-65 43-60 44 55 40-50 36 F-45, 32 40 28 35 24 F=30 20-25 16 20 26 8 F-10 4E5 o Eo et TIME in minutes g 333s 88 @ g egegaggsesess § 8 888888 8 8882 HE En bats + H Relative or actual Given any two corresponding quantities, solve > s CO LOCI 2/2 RBRVS rR 83 DISTANCE in miles for third by laying rule through points on prop- er scales and read intersection on third scale, F - OT ” = ry o nn @Oae 2 g g 8 a & 8 3 SPEED in knots MANEUVERING BOARD / Relative or actual Price 60 cents (per pad of 50) Washington, D.C., published May. 1920, at Hydrographic Office, under the outhority of the SECRETARY OF THE NAVY New Publication: 4th. Ed., Apr., 1938 4th. Edition, April 1938 No, 2065 a FIGURE 5. Case IV TO PROCEED FROM ONE RELATIVE POSITION TO ANOTHER AT MINIMUM SPEED GIVEN: COURSE AND SPEED OF GUIDE AND THE INITIAL AND FINAL RELATIVE POSITIONS FOR THE MANEUVERING UNIT. ; TO DETERMINE: COURSE AT MINIMUM SPEED AND TIME REQUIRED FOR MANEUVERING UNIT TO REACH FINAL POSITION. Example.—Guide on course 148° at speed 15.0 knots. A vessel, 7, now located 600 yards dead astern of the Guide, receives orders to open out to a distance of 19,000 yards and a bearing of 280° from the Guide, using minimum speed enroute - to conserve fuel. Required.—(a) Course and speed for M. (b) Time required to reach final position. (See fig. 6.) Procedure.—Plot Guide at G and the initial and final positions of M at M, and M. In this example, realizing that the Maneuvering Board is 20 divisions wide, a more accurate result is obtained by plotting G and M, on the final line of bearing and separated by 19 units. MW, is located 600 yards bearing 328° from G. Join M,.... Mp. From point e, lay out vectore. . . . gin direction 148°, length 15.0 knots. Transfer the slope M, . . . . M,to sg, extending it in the same direction from g that M, lies from M,. From e drop a perpendicular on this transferred slope, intersecting at m. ‘This is most readily done by observing the direction of the slope on the compass rose and subtracting or adding 90° to this. e. . . . mis the vector of the course and minimum speed for M. The time required to complete this evolution may be found by dividing the Relative Distance M, . . . . M2 by the Relative Speed g.... m. By using the Logarithmic Scale, this time may be obtained in minutes. Answer.—(a) Course 189° at minimum speed 11.4 knots. (b) 57 minutes. NOTE.—The foot of the perpendicular, m, must fall on the same side of g that Mz lies in respect to M;. Should it fall on the other side of g, it indicates that MZ must use a speed greater than that of the Guide and the solution becomes indeterminate. Fora plane, draw the wind vector,e . . . . w,and drop the perpendicular from w instead offrome. Thecorresponding air course is obtained by reference to w instead of e. Note that the course is always normal to the line of Relative Movement and can therefore be obtained without plotting if the direction of the latter is known. SCALES same aN are cual: SCALES 2:1 3:1 7. | Division=1000°Yards: , 4:1 5:1 Pes Ki | Division= ” 2. Knots “+ 76-95 36— 54 72—-99 u— SI 68 --85 32—4 48 64 --80 30—] 45 60 --75 28— 42 56 f-70 26— 39 52-65 24—| 36 48F-60 2543 44-55 ( 20— 30 40-50 tg—) 27- 36 F=45 6 24 32 F- 40 4 2i 28 E-35 2 18 24-30 1) 15 20F-25 8 2 16 20 6 9 Le ed) 4 6 8Eb 2 3 4ES5 o— 0 oo mS gags 8 3g 8 TIME In minutes g 88883 2 8gseeR DISTANCE In yards Relative or actual Given any two corresponding quantities, solve = me CS 1 eh) 2 R RRR’ 2 $8 DISTANCE In miles for third by laying rule through points on prop- er scales and read intersection on third scale. - nw ” z o o ne Doae = Rg a 8 ae g 3 SPEED in knots MANEUVERING BOARD L Relative of actusl Price 60 cents (per pad of 50) Washington, D.C., published May, 1920, at Hydrographic Office, under the authority of the SECRETARY OF THE NAVY Hew Publication: 4th. Ed, Apr., 1938 4th. Edition, April 1933 No. 2665 a FIGURE 6. 593046°—44_2 Case V TO PROCEED FROM ONE RELATIVE POSITION TO ANOTHER, REMAINING WITHIN GIVEN RANGE OF THE GUIDE FOR A SPECIFIED TIME ENROUTE GIVEN: COURSE AND SPEED OF THE GUIDE, INITIAL AND FINAL RELATIVE POSITIONS OF THE MANEUVERING UNIT, AND TIME TO REMAIN WITHIN A GIVEN RANGE OF THE GUIDE. TO DETERMINE: COURSE AND SPEED FOR MANEUVERING UNIT AND TRUE BEARINGS FROM GUIDE WHEN ENTERING AND LEAVING REQUIRED RANGE. Example.—Guide on course 200°, speed 15.0 knots. Ship M, now 35° on the starboard quarter of the Guide and distant 8.0 miles, receives orders take station 8.0 miles dead ahead of the Guide, remaining within 6,000 yards of the Guide for 45 minutes for the purpose of signalling. Required.—(a) Course and speed for M. (b) Time to come within required range and true bearing of Guide at that range. (c) True bearing of Guide when passing beyond required range. (d) Time to arrive at new station. (See fig. 7.) Procedure.—Plot Guide at G and locate initial and final positions of M at M, and M, respectively. M,, being 35° on the starboard quarter of G, bears 145° relative or 345° true from G. Join M,.... M2. About G draw a circle with radius of 6,000 yards (3.0 miles), intersecting M, .... M, at K and K’. Measure the distance K.... Ki. Lay oute.... gs, the vector of G. Transfer slope M, .... M.tog. Along this slope, from g, lay off the Relative Speed which is equal to the Relative Distance K . . . . K' divided by the time that M is to remain within the specified range. This is most readily done by utilizing the Logarithmic Scale, and thus m is located. Vectore ... . m indicates the course and speed for M to reach its new station while fulfilling the required conditions enroute. Distance M, .... K, divided by Relative Speed § . . . . m gives the time of arrival at the givenrange. M,....M, divided by § . . . . m gives the time to arrive on final station. Directions K....GandK!.... Gare the true bearings of the Guide from M when M reaches and passes beyond the given range respectively. Answer.—(a) Course 196°, speed 19.7 knots. (b) 73 minutes, bearing of Guide 12914°. (c) 056°. (d) 190 minutes or 3 hours and 10 minutes. NOTE.—Had the slope M, . . . . Mz been outside the required range, the obvious action indicated would have been to close to 6,000 yards in the most expeditious manner, take up the course and speed of the Guide for 45 minutes, and then proceed to the new station by another Maneu- vering Board set-up. 10 Given any two corresponding quantities, solve for third by laying rule through points on prop- @F scales and read intersection on third scale. MANEUVERING BOARD Price 60 cents (per pad of 50) ~, 1 Division=.1 Mite "”*. 1 Division? 3 Knots *. 70 65 55 45 25 200° 2 g gggeagheeses § 8 888388 - ; Relative or actual DISTANCE in miles 75 + 150 00d % DISTANCE in yards g 8 SPEED in knots Relative or actual Washington, D.C., published May, 1920, at Hydrographic Office, under the authority of the SECRETARY OF THE NAVY New Publication: 4th. Ed, Apr., 1938 4th. Editicn, April 1939 No. 2685 « FIGURE 7. ii Case VI (1) TO PASS AT SPECIFIED RANGE FROM GUIDE (SHOWN IN FIG. 8) (2) TO PASS OUTSIDE SPECIFIED RANGE OF GUIDE (SHOWN IN FIG. 9) (3) TO PASS WITHIN SPECIFIED RANGE OF GUIDE GIVEN: COURSE AND SPEED OF GUIDE, INITIAL RELATIVE POSITION OF MANEUVERING UNIT, COURSE OR SPEED REQUIREMENTS (IF ANY), AND RANGE (1) AT WHICH IT IS DESIRED TO PASS, (2) OUTSIDE OF WHICH TO PASS, OR (3) TO PASS WITHIN. TO DETERMINE: LIMITING COURSE, LIMITING SPEED, OR BOTH, FOR MANEUVERING UNIT AND TIME OF REACHING MINIMUM RANGE. Example A.—Guide on course 335° speed 18.0 knots. Ship M, now 18.0 miles bearing 340° from the Guide, wishes to pass the latter at a range of 8.0 miles. Required.—(a) Course or courses of MW at 12.0 knots if crossing ahead of G. (b) Course(s) of M at 12.0 knots if passing to eastward of G. (c) Speed of JV if course used is 295°. (d) Course(s) of M using minimum speed. (See fig. 8.) Procedure.—Plot Guide at any point G, and initial position of M at M;. About G draw a circle with radius equal to the given range. From M, draw tangents to circle about G, establishing points K and K’. M may either travel down the Relative Move- ment Lines M,.... KorM,. .. . K’, based upon either crossing ahead of and passing to westward of G or else passing to eastward of G. Transfer slopes M,....KandM,.. . .K’toS, cutting M’s 12.0-knot speed circle at m and mz for the former slope and at m3; and m, for the latter. Slope M, .... K also cuts the projected 295° course line at m;. WVectorse.... mu, e@.... M,C... . M3,0re.... m,can be used by M at the given speed. Vectore.... m; indicates M’s speed on the given course. To obtain minimum speeds and the courses corresponding, drop perpendicularse ....m,gande.... mm, fromeon the transferred slopes. Answer.—(a) 315° or 238°. (b) 347° or 100%°. (c) 10.0 knots. (d) Course 276!4° speed 9.5 knots or course 043}4° speed 6.5 knots. NOTE.—The vectors found above will cause M to pass G at exactly the given range. Ifit were desired to pass within the given range, the slopes M,.... Kand™M,. .. . K’ would be drawn so as to pass within the limiting range circle about G, and the same procedure followed. If it is desired to pass outside of the given range, the slopes M,. . . . Kand M,. .. . K’ and inclined so as to neither cut nor touch the limiting range circle, followed by the procedure mentioned above. A special application of this latter problem is given in example B on the following pages. It will be noted that where the speed of Mis less than that of the Guide, two solutions will be found for each transferred slope, unless the trans- ferred slopes are tangent to the speed circles, when only one solution is possible for each slope. In case the transferred slopes neither cut nor touch the speed circles, no solution is possible. The conditions imposed by the statement of the problem should indicate which of the resultant vectors to use. : 12 LE | Division =2 Miles | "+ .{ Division=3 Knots na 8 & & §ggeggeeeses 8 & 88egeeessse 8 Given any two corresponding quantities, solve a 20 Spel ne wm eolein es 1) 9 for third by laying rule through points on prop- er scales and read intersection on third scale. _ Relative cr actual 8 g DISTANCE in yards 8 DISTANCE in miles s J Ee on o 7 s © KR © oe 2 & yd illo ee ee eee SPEED in knots MANEUVERING BOARD iba Relative or actval Price 60 cents (per pad of 50) Washington, D.C., published May, 1920, at Hyd-tgraphic Office, > under the authority of tha SECRETARY OF THE NAVY © om Publication 40). Et, Ape, 1538 4D. Biitton, Apel 1903 No. 2085 a FIGURE &. 13 TO INSURE PASSING OUTSIDE OF A SPECIFIED RANGE FROM THE GUIDE WITH A COLUMN OF SHIPS, USING HEAD OF COLUMN. CHANGE OF COURSE Example B.—The leading vessel and flagship of a Cruiser Division of 3 ships in column formation, with a distance o 1,000 yards between vessels, is on Fleet Course of 135° at Fleet Speed 10.0 knots. Cruiser Flagship, now stationed 8,000 yard- broad on the port bow of the Fleet Guide, receives orders to proceed with the division to new station 8,000 yards on the stars board quarter of the Fleet Guide, keeping clear of the antisubmarine screen en route. The Cruiser Division Commander decides to use 20.0 knots, to allow 3,000 yards radius for the screen, and to make the change to new station by column move- ments, passing ahead of the Guide. Required.—(a) Course to cross ahead of the Guide. (b) Course to final station after clearing Guide. (c) Range and bearing of Fleet Guide from Cruiser Flagship when course is changed as in (b). (See fig. 9.) Procedure.—Plot Fleet Guide at G. Plot initial positions of the cruisers at CF, CA, and CB, CF being the Cruiser Flagship, and final position of CF at CF’. Draw Guide’s vectore.... &. Advance G along the Fleet Course the distance it would run while CB at 20.0 knots in advancing to the point where CF Speed of G 10.0 Soin Speed of CB of CB 2,000 yards or 20.0°°2” 000 yards=1,000 yards. This is done because in any column movement speed changes are made simultaneously while course changes are made in succession. About the advanced Guide’s position G’ draw a circle with radius equal to 3,000 yards, enclosing the antisubmarine area. From CF draw a tangent to this circle. From CF’ draw a second tangent to this offset circle, intersecting the first tangent at X. CF.... X represents the Relative Movement of the Cruiser Flagship in respect to the Guide while on the first leg. X.... CF’ represents similarly the Relative Movement while on the second leg. The bearing and distance of the Guide from CF at the turning point for starting the second leg is represented by X .... G. Transfer slopes CF .... Xand X.... CF’ to §, intersecting the 20.0-knot speed circle at c; and c, respectively. The vector of the first leg is e . . . . c, and of the second leg ise .... c&. Answer.—(a) 21112°. (b) 285°. (c) Range 4,200 yards, bearing 348°. NOTE.—This particular example is included to emphasize the fact that a column of ships may occupy considerable space. When maneuvering in the vicinity of other formations and units, all course and speed changes planned should take this entire space into consideration. Should the total length of the column be neglected, vessels following the column leader may be compelled to take an echelon formation to keep out of the danger area. This further reacts to increase the speed required to resume proper station in column, with consequent increase in fuel consumption. The pathCB....CA....Y.... CB’”, etc., indicates the Relative Movement of the last vessel in the column with respect to the Guide. While CB is advancing to the turning point for CF, its vector is indicated by e . . . . cb, and its speed relative to the Guide is § . . . . cb, or 10.0 knots. Using the Logarithmic Scale and the 3 minutes required for this advance, it is found that CB advances 1,000 yards relative to G, making it occupy the original relative position of CA, the changed to the first course. This distance is equal to second ship in column. From this point, its movement relative to G is indicated by the slope CA... . Y, which passes outside of the 3,000-yard antisubmarine screen circle about G. The Cruiser Flagship, as shown by its Relative Movement Lines CF ....X.... CF’, passes well outside of this 3,000-yard circle, but the extra distance travelled has prevented the embarrassing of the other ships in column and maintained better concentration. The positions occupied by the three ships on the first and second legs for any moment are shown by CF”, CA”, and CB” and by CF’’’, CA’’’, and CB’”’ respectively. 14 cee I Division=1000 Yards | Division= -” -2 Knots LOGARITHMIC SCALE TIME in minutes g gggsgge 8 & a 8 g g : 8 $8828R8 f FR pistance inyards F ca Relative or actual Given any two corresponding quantities, solve i be Gly iD On 2 gk SRees 2 8 DISTANCE in miles for third by laying rule through points on prop- er scales and read intersection on third scale. - ~ ” a ro) o Kr Oren 2g oe & a 8 3 ¢ 3 3 SPEED in knots MANEUVERING BOARD ies Ds Relative or actual * Washington, D.C., published May, 1920, at Hydrographic Office, Price 60 cents (per pad of 50) under the authority of the SECRETARY OF THE NAVY New Publicetion: 4th. Ed., Apr, 1938 4th. Edition, April 1938 No. 2665 < FIGURE 9. 15 Case VII WITH SPEED LESS THAN GUIDE, TO PASS CLEAR AT. MAXIMUM DISTANCE GIVEN: COURSE AND SPEED OF GUIDE, INITIAL RELATIVE POSITION AND SPEED OF MANEUVERING UNIT. TO DETERMINE: COURSE OF MANEUVERING UNIT AND MINIMUM RANGE ATTAINED. Example.—A submarine with maximum submerged speed of 8.0 knots wishes to keep clear as far as possible from a cruiser now bearing 135° distant 12.0 miles, whose course and speed have been determined as 345° and 24.0 knots. Required.—(a) Best course for submarine. (b) Minimum range and time to reach this range. (c) Relative bearing of cruiser from submarine at minimum range. (See fig. 10.) Procedure.—Plot the cruiser as Guide at G. Locate the present position of the submarine at S. Lay oute.... sg, representing course and speed of Guide. About e draw an 8.0-knot circle representing the maximum submerged speed of S. From gs draw tangents to this speed circle, points of tangency being s, ands. Vectorse....s,ande.... Ss; represent the two courses open to S at 8.0 knots. Transfer theslopesS ....s,andg....s8,toS. Although the Relative Speed is the same for either course it is readily apparent that the Relative Movement Line S.... X passes farther away from G than the line S....Y. @e.... is the best course for the submarine. Draw perpendiculars G....KandG....K’toslopesseS....XandS.... Y respectively G....Kis the minimum distance reached when on the course and speed represented by the vectore .... s,. The relative bearing is the difference between the bearing of G from K and the course indicated by vectore .... S. Answer.—(a) 274%°. (b) 9.15 miles; 20.8 minutes. (c) 180° relative or dead astern. NOTE.—If the slopes S.... XandS.... Y are on the same side of G, the preferable course under the required conditions is always away from the Guide. If vector e ... . s: is used, then the minimum range would be G ....K’ or 2.2 miles, with the cruiser bearing dead ahead of the submarine. G must be located between the slopes S ....XandS.... Y, for the submarine to make contact with the cruiser. 16 SCALES wi ae SCALES 2:1 3:1 _ 1 Division=.F Mile - . 4:1 5:1 38 57 «{ Division” 3 Knots *- 76 (95 36— 54 725-90 34 5! 68 F-85 32— 48 645-80 “30 45 60-75 28— 42 56 --70 26— 39 52 F-65 24— 36 48-60 22-5 33 445-55 20— 30 40F-50 18— 27 36-45 16— 24 32 F-40 4— 2l 28-35 12 18 24 F-30 0 6 20 F-25 8 2 16 --20 6 9 12 5 4 6 8 0 2 3 4 5 t) 0 0 0 4 8 2a8 & gs & TIME in minutes ess 8 8 8888s & 8 ns 8 & & gggeggs8ese8 & 8 SESE EREe fe eeess. § 8 pistance inyents : ; H Relative or actual Given any two corresponding quantities, sotve ze re CP SB ATCA =} 2 8 €RRVIS ® $8 DISTANCE in miles for third by laying rule through points on prop- er scales 4nd read intersection on third scale. = “ ” < eo © n@Oe 2 & & 8&8 8 ¢e gs 8 SPEED in knots MANEUVERING BOARD Relative or ectusl Price 60 cents (por pad of 50) Washington, D.C., published May, 1920, st Hydrographic Office, under the suthority of the SECRETARY OF THE RAVY Now Publication: 4th. Ed, Apr, 1938 4th. Edition, April 1938. No. 2865 a FIGURE 10. 17 Case VIII TO DETERMINE COURSE AND SPEED OF GUIDE FROM TWO SETS OF SIMULTANEOUSLY OBSERVED RANGES AND BEARINGS GIVEN: OWN COURSE AND SPEED, TWO SETS OF SIMULTANEOUSLY OBSERVED RANGES AND BEAR- INGS OF GUIDE FROM THE MANEUVERING UNIT, AND TIME INTERVAL BETWEEN OBSERVATIONS. TO DETERMINE: COURSE AND SPEED OF GUIDE, AND POSITION OF MANEUVERING UNIT RELATIVE TO GUIDE AT ANY SUBSEQUENT TIME. Example.—A destroyer rejoining a formation at a geographical point, after being on detached duty, is now on course 026°, speed 25.0 knots. The course and speed of the formation have not been signalled, since radio silence is in effect and the use of visual signals is restricted. At 1600 the first vessel in the formation is sighted and a simultaneous range of 29,000 yards and bearing of 007° obtained. This vessel is later identified as the Formation Guide, and at 1617 another observation is ob- tained, giving range of 20,000 yards and bearing of 014°. The commanding officer of the destroyer decides to maintain present course and speed until 1630, at which time he would head for this proper position in the formation. Required.—(a) Course and speed of the Guide. (b) Relative position of the destroyer at 1630. (See fig. 11.) Procedure.—Plot the Guide at G and the relative positions of the destroyer at 1600 and 1617 at D, and D, respectively. Measure the Relative Distance D, ....D. ~ Using the Logarithmic Scale, the time between observations, and the Relative Distance, determine the Relative Speed. Draw the destroyer’s vectore .... d, and transfer the slope D, .... D,tod. Using the Relative Speed found above measure it on this slope, locating §. e... . g is the vector representing the course and speed of the Guide. To locate the position of the destroyer relative to the Guide at 1630, continue the Relative Movement Line beyond D, and use the Logarithmic Scale. The estimated position D; may be plotted from either D, or D,, depending upon whether 30 or 13 minutes are used for length of time on Relative Movement Line. Answer.—(a) Course 066°, speed 14.7 knots. (b) 13,600 yards, bearing 206° from Guide. NOTE.—The vessel which is taking the bearings and ranges may be designated as the Guide in some problems, in which case it should be located first, and the other vessel plotted from the Guide’s position. The solution is similar. 18 "+ | Division=3000 Yards A (Division .° -3 Knots “-_ 76 56 52 24 TIME in minutes . DISTANCE in yards Relative or actual Given any two corresponding quantities, solve oS DISTANCE in miles for third by laying rule through points on prop- er scales and read intersection on third scale. 75 65 55. 45 35 20 MANEUVERING BOARD Tes pees , R (6) Relative or actual Price 60 cents (per pad of 50) Washington, D.C., published May, 1920, at Hydrographic Office, Ui - = ‘under the authority of the SECRETARY OF THE NAVY New Publication: 4th. Ed., Apr., 1938, 4th. Edition, April 1938 No. 2665 a FIGURE 11. 19 Case IX TO FIND THE COURSE AND RELATIVE POSITION OF A VESSEL FROM THREE OR MORE BEARINGS, KNOWING THE VESSEL’S SPEED GIVEN: OWN COURSE AND SPEED, SPEED OF SECOND VESSEL, AND THREE OR MORE BEARINGS WITH TIME INTERVAL BETWEEN BEARINGS. TO DETERMINE: COURSE OF SECOND VESSEL AND RELATIVE POSITION AT ANY SUBSEQUENT TIME. Example.—A ship A on course 350°, speed 14.0 knots, is taking radio bearings of ship B, which latter vessel is known to be making 10.0 knots. Bearings taken at 0813, 0858, and 0928, after being converted to true bearings, yield 017%°, 035°, and 050°, respectively. Required.—(a) Course of B. (b) Range and bearing of B from A at 1000. (See fig. 12.) Procedure.—Plot position of ranging ship A at any convenient point, and from this point lay out observed bearings AVA ee DingAe oh de: nD wanda, 4. fc tuwDas At any point lay out a slope across the bearing lines so inclined that intercepts between bearings are proportional to the time intervals between bearings. Methods of accomplishing this are shown on the following page. Letter this slope Ryeseeta, 2 Ro. oe: siecboas Lay out ranging ship’s vectore ....a. Transfer slope P; ....P,.... P3 to a, cutting 10.0-knot speed circle at b, and b,. ‘B therefore has two possible courses represented by vectorse ....b6,ande.... bp. Assuming B’s vector tobee .. . . b,, by using the time between the first and third bearings (75 minutes) and the Relative Speed a... . b, on the Logarithmic Scale, the Relative Distance B would run in this period of time is easily found. This Relative Distance B, .... B; is plotted between the 0813 and the 0928 bearings parallel to the slope P, .... P;. B; represents the position of B relative to A at 0928 under the assumption that e . . . . b, represents B’s vector. Similarly, if e .. .. by, is assumed to be proper vector for B, its relative position at 0928 plots at B’3. To find the Relative Positions possible for B at 1000, the respective Relative Speeds for the total time of 107 minutes on the Logarithmic Scale, give the Relative Distances run on courses shown by vectorse.... 6, ande.... bz locating B, and B’,. Answer.—(a) 108° or 001%°. (b) 39.4 or 8.8 miles, bearing 067°. NOTE.—When the speed of the second vessel is less than the speed of the ranging vessel, two courses will be obtained for the second vessel except when the transferred slope is tangent to the known speed circle. In case the bearing does not change, it will not be possible to determine the Relative Position of B by this method. All solutions obtained by bearings alone should be used with caution as a wide final error may result from a comparative small error in taking one or more bearings. This is borne out by the Radian Rule that ‘‘An error of 1° is an error of 1 mile at a distance of 60 miles.” 20 I Division 4 Miles . “+ «1 Division’*2 Knots *- £95 200° g 8 gesesseeess® # 8 88888888888 $F DISTANCE in ards - : Relative or actual $88 8 8 DISTANCE inmiles << 8 53 8 8 SPEED in knots Given any two corresponding quantities, solve for third by laying rule through points on prop- er scales and read intersection on third scale. MANEUVERING BOARD 1 i ne ges Price 60 cents (per pad of 5D) Washington, D.C., published May, 1920, at Hydrographic Office, under the authority of the SECRETARY OF THE NAVY Now Publicetion: 4th. Ed., Apr, 1928 4th, Edition, Aprii 1938 No. 2665 a FIGURE 12. 21 Techniques Used in Case IX TO DRAW A LINE ACROSS THREE BEARING LINES SO INCLINED THAT INTERCEPTS WILL BE PROPORTIONAL TO TIME INTERVALS BETWEEN BEARINGS Although a trial and error solution of this problem is possible by the use of parallel rulers and dividers, the three methods shown below have been found to be more expeditious. In each method the same data is used. Three bearings, O....a,O....b6,andO.... c, are taken by an observing vessel O of a second vessel, with an interval of time 1; between the first and second bearings and an interval of time ¢, between the second and third bearings. The angular difference between bearing O....aandO.... bis given by angle X, while angle Y indicates the difference between bearings O....bandO....c. These methods are shown sketched on a Maneuvering Board because of the convenience of the attached scales, but this is not a requirement. First Method (Sketch a). (See fig. 13.) At any point P on bearing line O . . . . a erect a perpendicular cutting bearing lineO ....batB. Along this perpen- dicular lay off B .... @so that ee ———— aoe Erect a perpendicular at Q, cutting O....catC. fe pkaalet ve 2 Connect B and C and extend this line to O .. . . a, intersecting at A. The required slopeisA....B....C. Second Method (Sketch bh). At any point P on bearing line O.... 5b erect a perpendicular and lay out P....@Q@ and P....R so that Se —— ==2 Erect perpendiculars at @ and R, cutting O.....aandO.... cat A and C respectively. flee ke 2 Connect A and C, cutting the bearing line O....b6 at B. Therequired slopeisA....B....C. Third Method (Sketch c). Along O.... a lay out any convenient distance O .... A. Along O....clay out length O.... C obtained by the formula: 6 RES GG ts Shi tag Snangle 2 t,; sin angle Y The required slopeis A....B... . C, obtained by connecting A and C. In case both angles X and Y are less than 16°, the numerical values may be used instead of the sines. TO DRAW A LINE OF GIVEN LENGTH PARALLEL TO ANOTHER LINE, BETWEEN: TWO DIVERGENT LINES (Sketch d) Let O....a,andO... . a: be two divergent lines and P....@Qagivenslope. It is required to draw a line of given length, parallel to P ....Q@, between O....a,andO.... a. Along P ... . Q@, extended if necessary, lay out the givenlengthP .....R. From Rdrawaline paralleltoO.... a, intersecting O..... a, at C. Through C draw A... . C parallel to P..... Q. A... . C is the required line, the proof of which is apparent from the parallelogram law. 22 16 27 24 2I MANEUVERING BOARD x Given any two corrésponding quantities, solve for third by laying rule through points on prop- er scales and read intersection on third scale. Price BO cents (per pad of 50) “(Sketch by -, Le SCALES 4:1 S: 76-95 72-90 68 Fes 64 80 60-75 56 F-70 52-65 48-60 . 44 55 (Sketch ce) 7 : : : pang Siggib Tab dagt anata ays ; ; pre Bs 40 so 36-45 s 32 40 28 F—-35 24 F-30 20F-25 16 F-20 12 5 8 10 yo Lah 4E5 (60 (Sketch d).*.’ . — 0&9 segs 8 g 8 TIME in minutes ggg 8 8 s 2 & 8 gggeggheseee 8 § Seese8 E8888 F 8 eee EE FF cistaNce inyant Relative or actual i SS oe Ss Ea 2 8 &€Sasze R$ DISTANCE in miles - rr) ” <= o o n @ae ae & a 85 ¢ 3 3 SPEED in knots I | f eee ee iii [yey Arendt frat bout J Relative or actual Washington, D.C., published May, 1920, at Hydrographic Office, under the authority of the SECRETARY OF THE NAVY Now Publication: 4th. Ed, Apr., 1938 4th. Edition, April 1933 No. 26654 FIGURE 13. 23 Case X TO DETERMINE SPEED OF GUIDE FROM THREE OR MORE BEARINGS, KNOWING GUIDE’S COURSE GIVEN: OWN COURSE AND SPEED, COURSE OF GUIDE AND THREE OR MORE BEARINGS OF GUIDE WITH TIME INTERVALS BETWEEN BEARINGS. TO DETERMINE: SPEED OF GUIDE AND RELATIVE POSITION AT ANY SUBSEQUENT TIME. Example.—Ship A, on course 140° at speed 12.0 knots, is taking bearings of ship M, which is known to be on course 180°. Bearings taken at 1130, 1230, and 1310 are 090°, 117°, and 14914° respectively. Required.—(a) Speed of M. (b) Estimated relative position of M at 1350. (See Fig. 14.) Procedure.—Plot the ranging vessel at any convenient point A and lay out bearings of M at 1130, 1230, and 1310 as Ae Ast epi, Aue) s) so, and Ale vo os 9ps,. respectively. By any of the methods described for case IX, draw slope P;. . . . P,.. . . P3 across the bearing lines at such an angle that the intercepts are proportional to the time intervals between bearings. From any convenient point e, lay out A’s vectore. . . . a, and M’sknowncourselinee.... m’. Transferthe slope P,.... P,... . P3;toa,cuttinge....m’atm. ee... . mis the vector representing the course and speed of M. Using the Relative Speed indicated by vectora . . . . m, the Relative Distance run by M during the time the bearings were taken is found by means of the Logarithmic Scale. By trial and error or the method indicated in case IX, this is plotted, locating M,, M2, and M;. These positions are those occupied by M relative to A at 1130, 1230, and 1310, respectively. The relative position of M at 1350 is also found through the Logarithmic Scale as a 2"20™ run from 1130 or a 40-minute run from 1310, and is designated as M,. Answer.—(a) Speed 14.5 knots. (b) Bearing 167°, distant 10.7 miles. NOTE.—This case becomes indeterminate if the slope P} . . . . Po . . . . P3is parallel to the vector of the ranging ship. This can occur only if M is on the same course as A or the reverse of this course. In this event, since the course of M is known, A should steer another course for taking bearings. If the bearing does not change, it will not be possible to determine the relative position of M by this method. 24: SCALES 2 38 36 32 28 26 24 20 18 rto3:1 57: SI 45 42 39 36 27 24 2i LOGARITHMIC SCALE Given any two corresponding quantities, solve for third by laying rule through points on prop- er scales and read intersection on third scale. MANEUVERING BOARD Price 60 cents (per pad of 50) 593046°— 44 ___3 25 Washington, D.C., published May, 1920, at Hydrographic Office, under the authority of the SECRETARY OF THE NAVY FIGURE 14. New Publication: 4th. Ed. Apr., 1938 | Division=2 Miles: 4tn. Edition, April 1938 TIME in minutes DISTANCE in yards Relative or actual DISTANCE in miles SPEED in knots Relative or actual No. 2605 a 25 : Case XI TO DETERMINE COURSE, SPEED, AND RELATIVE POSITION OF A TARGET BY BEARINGS ALONE GIVEN: TWO SETS OF THREE OR MORE TIMED BEARINGS, EACH TAKEN ON A TARGET BY AN OB- SERVING UNIT WHICH CHANGES ITS OWN KNOWN COURSE OR SPEED OR BOTH BETWEEN SETS OF BEARINGS. TO DETERMINE: COURSE, SPEED, AND RELATIVE POSITION OF THE TARGET. Example.—An observing vessel G, while on course 110°, speed 15.0 knots, obtains radio bearings on target vessel M as follows: Time 0800. Bearing of M 000°. Time 0900. Bearing of M 349°. Time 1000. Bearing of M 336°. At 1015 the observing vessel changes course to 045° and increases speed to 20.0 knots. Bearings are next received as follows: Time 1030. Bearing of M 327°. Time 1130. Bearing of M 302°. Time 1230. Bearing of M 273°. Required.—(a) Course and speed of M. (b) Relative position of M at 1230. (See fig. 15.) Procedure.—Plot position of observing ship at any convenient point G, and lay out the 0800 bearing line as G.... by, the 0900 bearing asG. . b2, and the 1000 bearing as G . . b3. By any of the methods shown for case IX, draw a slope P eesdcn@s: . R across these bearing lines so inclined that! the intercepts P . . @and @.... R are proportional to the time retails facewieeri bearings. In a similar manner lay out the second set of bearings, G....bi G....b6;, and G....b5,;. Draw the slope T....U.... Vso inclined that the intercepts T....UandU.... V are proportional to the time between these bearings. From any point e, lay out the first vector of G ase.... g:, and transfer the slope P....@....Rtog,. Draw the second vector of G,e.... g, and transfer the slope T....U....Vtog:. This intercepts the slope from g; at m. e....mrepresents the course and speed of M. Determine the Relative Speed s,....m and by means of the Logarithmic Scale find the Relative Distance travelled by M between the 1030 and the 1230 bearings. Lay off this distance, T....W, and by completion of the parallelogram locate the position M at 1230. T’....M is equal to T....W £ and is the Line of Relative Movement between 1030and 1230. M’s bearing and distance from G at 1230 is G....M. Answer.—(a) Course 164°, speed 12.6 knots. (b) 57.5 miles bearing 273° from observing vessel. NOTE.—Solution by this method will not be obtained if the second vector of G should lie along the transferred slope P....Q@....R. For this reason this slope is transferred to g, before G changes either course or speed or both. If the bearing does not change in either set, a solution is still obtainable. The slope in this case is a constant bearing and is laid off in both directions from the head of the vector concerned. In case G makes a change of course when the last bearing of the first set is obtained, this beans may be used as the first bearing of the second set. When bearings alone are available, the results should be considered as approximations only. This is occasioned by the fact that a small error in one or more bearings will change the inclination of the slopes to be transferred, and this in turn will change the final results. If sufficient time is available, a third set of bearings will act to check the course and speed of the target. 26 _ 1 Division=19'Milés . "+ | Division = -2 Knots we Stk t + 1 ao } Boot a al f ec)! | leet: gt | se t | he LOGARITHMIC SCALE gggsss § 8 889888 3 4 8 8 8 § 8 8 g g2ga8 82828 8 g 8 28288 8 8888 8 $ 882882 (8° 8 pistance inyards : r 7 Relative or actual Given any two corresponding quantities, solve ‘a = OS OF “ass pO ets 2 RRRALSS 2 8 DISTANCE in miles for third by laying rule through points on prop- er scales and read intersection on third scale. : . , 2 0 ” < o © n eae 2 & a 8: 2 & 8 SPEED in knots MANEUVERING BOARD : Relative or actual Price 60 cents (per pad of 50) ; Washington, .C., published May, 1920, at Hydrographic Office, under the authority of the SECRETARY OF THE NAVY New Publication: 4th. Ed., Apr., 1938 4th. Edition, April 1938 No. 2665 a FIGURE 15. 27 Case XII TO OPEN TO GIVEN RANGE IN MINIMUM TIME GIVEN: COURSE AND SPEED. OF GUIDE, INITIAL RELATIVE POSITION, OWN SPEED AVAILABLE, AND RANGE TO BE ATTAINED IN MINIMUM TIME. TO DETERMINE: COURSE OF MANEUVERING UNIT AND TIME TO REACH SPECIFIED RANGE. Example.—Guide on course 300°, speed 14.0 knots, has vessel M now bearing 000° and distant 15,000 yards. .M, which _ has power available for 21.0 knots, receives orders to open out to 30,000 yards from the Guide as quickly as possible. Required.—(a) Course for M. (b) Time required for M to reach the given range. (c) Relative bearing of M from G when given range is reached. (See fig. 16.) a Procedure.—Plot Guide at any point G, and locate initial position of Maneuvering Unit at M. About G draw a circle with a radius of 30,000 yards. Speed of G Speed of M@ From X, draw a line through M, intersecting the 30,000-yard circle at Y. M.... Y is the Relative Movement Line of M. The explanation for the formula is appended below. Transfer the slope X....M.,... Yto gin the Vector Diagram, intersecting the 21.0 knot speed circle at m. e....m represents the course of M at 21.0 knots to open to the required range in minimum time. The time required is found from the Logarithmic Scale, using Relative Distance M.... Y and Relative Speed §....m. Bearing of M from G at 30,000 yardsisG.... Y. Answer.—(a) 046°. (6) 23.7 minutes. (c) 106° relative. NOTE.—Should the speed of G be greater than the speed available to M, the point X will lie outside the specified range circleand X.... M, as extended, will cut the range circle twice. The point Y should be so chosen, in this case, that M lies between points X and Y. From G, in the direction of the Guide’s course, lay out G.... X equal to 30,000 yards=20,000 yards. Explanation.—The geometric construction used in the solution of this problem is based on the fact that the shortest distance from a point within a circle to its circumference is along a radius passing through that point... Therefore, if M is to reach the 30,000-yard circle from G in the minimum time, M must run along the radius of a circle whose center is the navi- ational position of G at the instant the circumference is reached by M. Consider that the time required for this evolution is ¢ hours. In that length of time the Guide will have traveled 14¢ miles, plotted as the distance G.... G’, along course 300°. From G’ draw a line through M equal in length to the specified range. This lineis G’....M....A, and M.... Ais equal to the distance traveled by M in ¢ hours or 21¢ miles. Com- plete the parallelogram whose adjacent sides are G....G’ and G’....M....A. It will readily be seen that the triangles XGY, XG’M, and YAM are similar since their sides are parallel. The proportionality follows: (Ag Se SORA Sas o (DEO oe 8 OK 6 de 2) ™ (GE) (G. G’) onl(G i2h). O= Grex -Y)X or... A) =" a ese 4) XM... A) A) Now G.... Y is equal to the limiting distance and G....G’ and MW....A are drawn equal respectively to 14¢ and Pee : ; Abus : 14¢ Therefore G:... X=limiting distance Xoi¢ speed of Guide speed of Maneuvering Unit™ This problem is possible.of solution by the trial and error method but much laborious work may be avoided by the geo- metric method described above. =limiting distance X 28 "+" | Division +3000 Yards 3 | Division= .° 3 Knots “+. TIME in minutes ; gf 8 ggggess & 8 8 3 8 g g 8 g gagge 28888 8 8 8 8 e883 8 888) 8 $ B2seRgRs 2 8 DISTANCE in yards ao 1 Relative or actual Given any two corresponding quantities, solve - ” do OT NE 2 RRRRLIS ® 8 DISTANCE in miles for third by laying rule through points on prop es scales and read intersection on third scale. - 7] ” <= o o nm @oaec ~) & 2 83 2 ¢ 3 3 SPEED in knots MANEUVERING BOARD fl Relative or actual ‘i Washington, D.C., published May, 1920, at Hydrographic Office, Price GO cents (per pad of 90) under the authority of the SECRETARY OF THE NAVY New Publiction: 4th. Ed, Aor, 1933 4th. Baition, April 1938 No. 2665 a FIGURE 16. 29 Case XIII GIVEN: COURSE AND SPEED OF GUIDE, INITIAL POSITION AND SPEED RESTRICTIONS OF MANEU- VERING UNIT, AND RANGE TO BE REACHED IN MINIMUM TIME. TO DETERMINE: COURSE OF MANEUVERING UNIT AND TIME TO REACH SPECIFIED RANGE. Example.—A destroyer now stationed 10.0 miles 80° on the starboard bow of the Fleet Guide receives orders to close to 4,000 yards from the Guide as quickly as possible. The destroyer is capable of making 24.0 knots. Fleet course is 020°; fleet speed 15.0 knots. Required.—(a) Course of the destroyer. (b) Time to reach 4,000-yard (2.0 mile) range. (c) Relative bearing of D from G when range is reached. (See fig. 17.) Procedure.—Plot the Guide at any convenient point G, and locate the initial position of the destroyer, bearing 080° relative or 100° true from G. Mark this point D. From G, in a direction the reverse of the Guide’s course, lay out G. .. . X equal to sees of C Range to be reached or 1.25 miles. About G draw a circle with 2.0 mile radius. Join D.... X, cutting the 2.0 mile circle at Y. D.... Yis the line of Relative Movement of D. Lay out the Guide’s vectore .... g. Transfer the slopeD....Y.... X to S, intersecting the 24.0-knot speed circle at d. e.... d represents the course and speed of D. The time to arrive at the specified range is found from the Logarithmic Scale, using the Relative Distance D.... Y and the Relative Speed $ .... d. Answer.—(a) Course 310°. . (b) 21.3 minutes. (c) 110° relative from G. NOTE.—In case the speed of D is equal to the speed of G, the point X will lie in the given range circle. If the speed of D is less than the speed of G, the point X will lie outside the circle, and the possibility exists of D . . . . X intersecting the range circle twice. In this event, theinter- section nearer to D is the one to be used as Y. In the case the initial position of the Maneuvering Unit is such that the line D . . . . X does not intercept the range circle, then the problem is incapable of solution. 30 I Division=.f°Mile 4:1 5:1 “A Division=3 Knots ee 76-95 725-90 68 F-85 645-80 605-75 56-70 52 --65 485-60 445-55 405-50 36 F—-45 32 F-40 28 F-35 24-30 20-25 16 au 12 15 8 0 4 5 0 i) @s3 8 g§ 8 TIME in minutes : gggees 8 8 & § $8 gees 2988 o q 33288 8 & pstance In yards Reistive or actual Given any two corresponding quaniities, solve F J CPAP NOT Sd J 2 § ASRS R $8 DISTANCE In miles for third by laying rule through points on prop- er scales.and read intersection on third scale. = ~ ” = » © mn @ae » & eset 3 & 8 SPEED in knots MANEUVERING BOARD Relative or actual ; 4 .C., published May, 1920, at phic Office, Price 60 cents (per pad of 50) mes cbt a eS ie New Publication: 4th. Ed, Apr, 1938 4th. Edition, April 1938 No, 2665 @ FIGURE 17. 31 Case XIV TO REMAIN WITHIN: SPECIFIED RANGE FOR MAXIMUM TIME - § GIVEN: COURSE AND.SPEED OF GUIDE, INITIAL RELATIVE POSITION AND SPEED OF MANEUVER- ING UNIT, AND RANGE WITHIN WHICH IT IS DESIRED TO REMAIN FOR THE MAXIMUM TIME. TO DETERMINE: COURSE OF MANEUVERING UNIT AND LENGTH OF TIME MANEUVERING UNIT REMAINS WITHIN THE REQUIRED RANGE. iA Example.—A cruiser is proceeding.on course 190° at 21.0 knots. A merchant vessel, now bearing 280° and distant 5.0 miles from ‘the cruiser desires to remain within 10.0 miles of the cruiser as long as possible; maintaining its best speed of 12.0 knots. _ AG. Required.—(a) Course for the merchant vessel. (b) Length of time remaining within specified range. _(c) Relative bearing of the cruiser when the limiting range is reached. (d) Minimum range reached. (See fig. 18.) Procedure.—With the cruiser as Guide, plot its position at any convenient point G. Plot the merchant vessel’s initial position at M and about G draw a circle of radius 10.0 miles. iapler : ‘ d Speed ae Lay off G.. . . . Xin direction the reverse of the cruiser’s course and in length equal to Speed atiG s Specified range. Speed of M Join M and X, cutting the 10.0 mile circleat Y. M.... Yis the line of Relative Movement for M while within the required range. ; Lay out G’s vector,e .... g, andtransfer the slope M....Y.... Xto g, cutting the 12.0 knot speed circle at m. Vector’e .... msrepresents the course and speed for M. By means of the Logarithmic Scale, the Relative Distance M .... Y, and the Relative Speed g . . . . m, the length of time that the merchant ship remains within the required range is readily found. A perpendicular from the line of Relative Movement through the position of the Guide indicates the closest approach to the Guide. Answer.—(a) Course 177°. (b) 63 minutes. (c) Dead ahead. (d) 4.8 miles. NOTE.—If the speed of M is equal to the speed of G, the point X will fall on the given range circle, locating Y at this point. The course of M will then be parallel to the course of G, and since both are making the same speed, the initial relative positions will be maintained. Ifthe speed of Mis greater than the speed of G, the line G . . . . X should be plotted in the same direction as the Guide’s course and the point Y located by extending the line G ... . X until it cuts the required range circle. When the required range is reached, in this event, the Guide will bear dead astern of the Maneuvering Unit. 32 | Division'=2 Miles . “+ 1 Division=3 Knot 83a § 8 : 8 8 #888888 8 & a 8 8 @ 8 ¢gggegseesss 8 & 8 88288 82888 : ee eee 3 8 Given any twe corresponding quantities, solve > ee GC) St Dhe oo 2 RR S828 2 8 for third by laying rule through points on prop- er scales and read intersection on third scale. - c) i) 2 rv) on Dae = 8 2 es 83 ¢ gz 3 MANEUVERING BOARD Price 60 cents (per pad of 5D) Weshington, D.C., published May, 1920, at Hydrographic Office, tnder the euthority of the SECRETARY OF THE NAVY New Publication: 4th. Ed, Apr., 1933 4th. Edition, April 1938. FIGURE 18. TIME in minutes DISTANCE In yards Relative or actual DISTANCE in miles SPEED in knots Relative or actual No. 2665 a 33 Case XV TO REMAIN OUTSIDE A SPECIFIED RANGE FOR MAXIMUM TIME GIVEN: COURSE AND SPEED OF GUIDE, INITIAL RELATIVE POSITION AND SPEED OF MANEUVER- ING UNIT, AND RANGE OUTSIDE OF WHICH IT IS DESIRED TO REMAIN FOR THE MAXIMUM TIME. TO DETERMINE: COURSE OF MANEUVERING UNIT AND RESULTANT TIME MANEUVERING UNIT REMAINS OUTSIDE OF THE SPECIFIED RANGE Example,—A cruiser on course 210° at a speed of 18.0 knots orders a tanker, now located 10° on the port bow and distant 14.0 miles, to remain outside of a range of 10.0 miles from the cruiser as long as possible. The tanker can make 10.0 knots. Required.—(a) Course for the tanker. (b) Length of time tanker remains outside the specified range. (c) Relative bearing of the cruiser when the 10.0-mile range is reached. (See fig. 19.) Procedure.—Plot the position of the cruiser at any convenient point, G, and the initial position of the tanker, the Maneu- vering Unit, at M. Draw the 10.0-mile-range circle about G. Speed of G From G, lay off the line G .... X, in the same direction as the course of G and in length equal to Speed of M*< Specified range. Join X and M, extending the line until it cuts the 10.0-mile-range circle. It will be noted that this line inter - sects the circle at two points, Y and Y’. The first intersection, Y, is the one to be used as it is nearestto X. M....Y is the line of Relative Movement. Lay out.the Guide’s vectore .... g, and then transfer the slope MM... . Y to g, cutting the 10.0-knot-speed circle at mand m/’. The course for the Maneuvering Unit is indicated bye .... m. By means of the Relative Distance MM... .Y and the Relative Speed § . . . ..m, the time that the Maneuvering Unit will remain outside of the 10.0-mile range is readily found from the Logarithmic Scale. The true bearing of G from M at the time that the 10.0-mile range is reached is given by the line Y . . . . G, which is the reverse of the course of the Maneuvering Unit. Answer.—(a) Course 175°. (b) 35 minutes. (c) Dead astern. NOTE.—Unless the initial position of M lies within the area bounded by the tangents from X to the given range circle, and the arc of this circle between the points of tangency, the Maneuvering Unit need not worry about coming within the prescribed range. If the initial position is eutside of this area, the Maneuvering Unit can remain outside of the range indefinitely. A glance at the Vector Diagram will show why the vector e . . . . m and not the vector e .... m’ was used for the Maneuvering Unit. While the Relative Distance remains the same before the range is reached, the Relative Speed would be increased from g....mtog .... m/’ and the time remaining before reaching the range thereby reduced. This is contrary to the results desired. ___ ( Division=2 Miles , 4:1 5:1 “+1 Division=3 Knots *- 76 95 72 F-90 68 F-85 64F-80 60-75 56 [-70 525-65 48-60 445-55 40-50 36 F—-45 32 40 23 F—-35 24 F-30 20-25 16 --20 2-5 8 10 465 i) 0 ae TIME in minutes se f gegsgee 8 & 2 8 8 @ 8 sgegaggseesss 8 £8 sees 88888 8 882888 8 2B pistance inyards AF Relative or actual Given any two corresponding quantities, solve is vw es yererg £2 & RF8S3SZ RK B__ DISTANCE inmiles for third by laying rule through points on prop- er scales and read intersection on third scale. = n ” < o o nn @ae 2 g 2 2 8 2 3 3 SPEED in knots MANEUVERING BOARD Relative or actual i Washington, D.C., published May, 1920, at Hydrographic Office, orem GD iesrda) (por val) Ae the authority of the SECRETARY OF THE NAVY New Publication: 4th. Ed., Apr, 1938 4h. Edition, April 1938 No, 2665 a FIGURE 19. 35 Case XVI TO SCOUT FROM INITIAL RELATIVE POSITION IN A GIVEN GENERAL DIRECTION AND RETURNING TO A FINAL RELATIVE POSITION, SPEEDS AND TIMES ON BOTH LEGS BEING SPECIFIED GIVEN: COURSE AND SPEED OF GUIDE, INITIAL.AND FINAL RELATIVE POSITION OF SCOUT, GEN- ERAL DIRECTION OF FIRST COURSE, AND TIME AND SPEED OF SCOUT ON EACH LEG. TO DETERMINE: COURSES OF SCOUT AND RELATIVE POSITION AT TURNING POINT. Example.—Guide is on course 320°, speed 12.0 knots, with scout A now stationed 10.0 miles broad on the starboard beam of the Guide. A is ordered to scout in a northerly direction, using a speed of 20.0 knots for 4 hours and a speed of 15.0 knots for 4 hours. At the end of the 8 hours, A must reach its new station, 10.0 miles on the starboard quarter of the Guide, ready to assume course and speed of the Guide. Required.—(a) Course of A on first leg... (b) Course of A on second leg. (c) Relative position of A from the Guide at turning point. (See fig. 20.) Procedure.—Since the Guide is on a steady course for 8 hours and the distances are fixed that A must travel on each leg; it is readily apparent that regular chart work, or the Navigational Plot, is indicated. Plot the initial position of the Guide and of A at Gj and A, respectively. Advance the position of the Guide to G, for the run of 8 hours or 96.0 miles on course 320°. Locate A;, the position of the scout at the end of 8 hours. With A, as a center, draw a circle with radius of 80.0 miles, the distance covered by the scout on the first leg. Similarly, with A, as a center, draw a circle with radius of 60.0 miles. These circles intersect at K and K’. By the conditions of the problem A is to scout in a northerly direction, so A, ... . . K represents the course on the first legand K . . .. A, the course on the second leg. To find the relative position of A at the turning point, locate G’, the position of the Guide at the end of 4hours. G’.... K represents the bearing and distance of A from the Guide. Answer.—(a) 359°. (b) 257°. (c) 62.5 miles bearing 037° from the Guide. NOTE.—In case the initial course for A had been specified instead of the initial speed, this course would have been laid out from Aj, inter- secting the circle about Az: at X and X’. The courses for the second leg, in this case, would be X .... A, or X’ .... As, depending upon whether the speed indicated by A, . .... X divided by 4 hours or A; ... . X’ divided by 4 hours is used. 36 SCALES wt wy Spisarentes . 80 : Dione ; : 2:1 3:1 ; peeoe ge : 3 sees | Division=10 Miles , 38— 57 ah y : : . eg ao—| 54 34 SI 32— 48 30—] 45 28— 42 26 39 24-4 36 2133 20-30 1s 27 te] 24 u—| 2 2—| 13 lo 15 8-2 6—| 9: 4-16 2 3 oto TIME in minutes g 8 a 3 g 8 88 geged 28398 8 8 a gages $8888 2 5 aggeee 2 g DISTANCE in yards Given any two corresponding quantities, solve y = a ae ead 2 a2 8 RSS28 2 8 Gere Wines for third by laying rule through points on prop- er scales end read intersection on third scale. a ov ml ge hn ako Bm oORe =) Be Wa eeieeks Be hs SPEED linlieuts MANEUVERING BOARD psy | Relative oF actual Price 60 cents (per pad of 50) Washington, D.C., published biay, 1920, at Hydrographic Office, under the authority of the SECRETARY OF THE NAVY Hee Publication: 4th. Ed, Apr, 1933 401 Edition, April 1938 ‘No. 2665 a. FIGURE 20. 37 Case XVII-A LOCUS OF MEETING POINTS OF TWO SHIPS GIVEN: INITIAL RELATIVE POSITION AND SPEED OF EACH UNIT. TO DETERMINE: AREA WITHIN WHICH THE TWO UNITS CAN MEET. Example.—A ship §, with 10.0 knots speed, is now located 60.0 miles bearing 180° from ship F, which has 20.0 knots speed. Required.—(a) Locus of meeting points for F and S, using given speeds. (b) Limiting courses open to F at its given speed. (c) Corresponding courses for S at its given speed. (See fig. 21.) Procedure.—Plot the slower ship at any convenient place S and locate the faster ship at F. Extend the line F ....S indefinitely. The two ships will meet earliest when they are headed directly for each other, for then the Relative Distance is being traveled at their combined speeds. The latest time of meeting will similarly occur when both ships are on the same course and the faster ship is headed directly for the slower ship, for then the Relative Speed is the difference between the two speeds. The points of earliest meeting and of latest meeting are conveniently located by travel of the slower vessel since the distances traveled are less. The point of earliest meeting, P;, and the point of latest meeting, P:, are the extremities of the diameter of a circle whose circumference marks the locus of positions at which the two vessels can meet when using the given speeds. In case the slower vessel makes less than its given speed, the area enclosed by this circle represents the locus of meeting places. __ Original separation X Speed of S, By means of the formula, S .... Pi= Succtasi@nhisiSaecdionS locate P;. In this particular instance, both the time of earliest meeting and the distance S . . . . P; may be found by using the Logarithmic Scale. Original separation X Speed of S SS... . P, is then equal to Speed of F minus Speed of S - The center of the circle, O, may be found by bisecting the line P; . . . . P2: or by the formula, Seb fia oases: Fi piss Se Pi minus S Ce Pa Tangents drawn from F to the locus circle drawn from O with radius O . . . . P, are the limiting courses that F may take when both ships use their given speeds. Corresponding courses for S are from S to the points of tangency, or S .. . . K’ and S.... K respectively. Answer.—(a) Circumference of a circle of radius 40.0 miles and center 20.0 miles bearing 180° from S. (b). Courses be- tween 150° and 210°, measured clockwise. (c) 090° if F’s course is 150°; 270° if F’s course is 210°. NOTE.—In case the speed of F is equal to the speed of S, the radius of the locus circle becomes infinite, and a perpendicular erected at the midpoint of S . . . . F becomes the required locus. In the event that F decides to steer a course between the tangents, such as F .... X’, then S has a choice of 2 courses, S .... X and S.... 2’. Inthe event that S decides to steer some intermediate course such as S .... Y while F is moving along the line F .. . . X’, the time of meeting will be found by dividing the distance F .... Y by F’sspeed. S will then use a speed lower than 10.0 knots, found by dividing the distance S .... Y by the time previously found. Once the required locus has been found, one ship decides on the course she is to steer and so notifies the other vessel, which must set her course to reach that meeting point. 38 tdci lelelaletnat-lets eller rat net NAVIGATIONAL PLO Ein minutes 8 3 = ® DISTANCE in yards am Relative or actual Given any two corresponding quantities, solve = R ® 8 _ DISTANCE in miles for third by laying rule through points on prop- er scales and read intersection on third scale. 7 ~ Ly? * Ge Sa er SPEED in knots MANEUVERING BOARD Relative or actual Price 60 cents (per pad of 50) Washington, D.C., published May, 1920, at Hydrographic Office, under the authority of the SECRETARY OF THE NAVY Now Publication: 4th. Ed, Apr., 1938 4th. Edition, April 1938 No, 2665 a. FIGURE 21. 39 Case XVII-B MEETING POINTS OF THREE SHIPS AT GIVEN SPEEDS GIVEN: POSITIONS AND SPEEDS OF THREE SHIPS THAT ARE TO MEET SIMULTANEOUSLY. TO DETERMINE: MEETING POINTS, COURSE FOR EACH SHIP, AND TIME OF MEETING. Example.—Ship A has ship B bearing 120°, distant’ 40.0 miles, and ship C bearing 030°, distant 25.0 miles. ‘Speeds available to the ships are as follows: A, 18.0 knots; B, 12.0 knots; and C, 16.0 knots. It is required that the three ships meet simultaneously at their best speeds. ke oe Required.—(a) Location of earliest-meeting point relative to A. (b) Course for each ship to earliest meeting place. (c) Time for-éarliest meeting. (See fig. 22.) Procedure.—Plot the positions of the three ships at A; B, and C. By application of case XVII-A, determine the locus of meeting points of ships A and B, A and C, as well as ships B and C. These loci intersect at K and K’. K, being closer to the initial positions of the three ships, is the earliest meeting point. Time for earliest meeting is found by dividing A ....K by speedof A,B.... clan FMA 2 F RPSRR HR F "Mond nl EEER2. SR $8 RB oR oe = Fecapisbe sted nae = bi 2 1 5 : ; RoMHO aitegergoniyt fe LIT Spe odtelhse,.0.6 pe 200E oh SEOL QA wot OR ERE PF IH stan wo WANBNT 20 YRATZTORS wi fe wherthe at mew BS SAHUDIT SECTION II TWO COURSES FOR GUIDE, SINGLE, COURSE FOR MANEUVERING UNIT | | The cases thus far discussed have comprised either those in which both Guide and Maneuvering Unit maintain constant | lcourses and speeds throughout the problem, or else they have involved successive applications of cases of this type» In this section we deal with situations which require a change of course or:speed.or both on the part of the Guide while the Maneu- vering Unit is proceeding from one Relative Position to another. When the total time for the maneuver is given, as well as the exact time the Guide changes course or speed or both, the obyious and most rapid solution is a simple Navigational Plot. When the Navigational Plot may not be thus employed, the problem may be solved by so manipulating or distorting the Relative Movement Diagram that. the cases resolve themselves into the Single Vector Cases discussed in section J. This idistortion is required because the Guide does not maintain its course and speed throughout the problem, and consists in em-~ jploying an imaginary or Fictitious Guide which does maintain its course and speed throughout the problem. The use of a ictitious Guide should be readily understood by any practical navigator who has had to compute the net run of a vessel in Traverse Sailing, for finding the Course and Distance made good between Fixes of Position. This Fictitious Guide may be considered as accompanying the real or actual Guide on either its first or its second leg, idepending upon the conditions of the problem., ‘The course and speed of the Fictitious Guide will therefore be represented iby one of the vectors of the real.Guide. Since the Fictitious Guide uses this vector throughout the problem, it may be em- Iployed in exactly the same manner as previously described in section I. In the Relative Plot, positions are located with reference to the real Guide by the statement of the problem. If, however, |the problem requires the employment of a Fictitious Guide for purposes of solution, it is essential that all positions in the |Relative Plot be referred to. the Fictitious Guide (reorienting as necessary) and not to the ae Guide. As long as the real Guide jand the Fictitious.Guide are together, any position plotted relative to the one is plotted relative to the other; but this relation- {ship does not hold when the real Guide and the Fictitious Guide are not together. In order that we may een our. Fictitious |Guide stationary in the Relative Plot, itis necessary to offset one of the positions of the Maneuvering Unit by the amount |this position would be shifted with respect. to the Fictitious Guide during the time that the Fictitious Guide and the real }Guide are not in company. In general, the initial set up of the Vector Diagram comprises the vectors representing the two courses and speeds of the jactual Guide, one of which vectors will therefore also be the vector of the Fictitious Guide. The Relative Plot initial set up jconsists of the real Guide and the Fictitious. Guide located together at any convenient point, and the initial and the final \relative positions of the Maneuvering Unit plotted with reference to the real Guide. One of these Relative Positions is next Joffset the proper amount to relocate it in respect to the Fictitious Guide. .The other relative position remains untouched jsince it is plotted for the time when the real Guide and the Fictitious Guide are together. The following rules govern cases involving employment of the Fictitious Guide: (1), When the time of departure of the Maneuvering Unit is known, the vector of the Fictitious Guide coincides with the second vector of the real Guide. The initial relative position of the Maneuvering Unit is therefore offset by the relative run \of the real Guide with respect to the Fictitious Guide, while the real Guide is on its Jirst course. (2) When the time of arrival of the Maneuvering Unit at its final relative position is known, the vector of the Fictitious | Guide coincides with the first vector of the real Guide. The final relative position of the IVinmeeseee Unit is therefore offset | by the relative run of the real Guide with respect to the Fictitious Guide, while the real Guide is on its second course. (3) The direction of offset is always toward the second vector of the real Guide. 55 Case XX TO PROCEED FROM ONE RELATIVE POSITION TO ANOTHER AT KNOWN SPEED, ARRIVING AT GIVEN TIME, GUIDE CHANGING COURSE DURING OPERATION GIVEN: COURSES ‘AND SPEEDS OF GUIDE AND TIME AT WHICH GUIDE WILL CHANGE COURSE, INITIAL/-AND FINAL RELATIVE POSITIONS, SPEED OF MANEUVERING eles AND TIME OF ARRIVAL AT FINAL POSITION. 2 ¥ TO DETERMINE: COURSE OF MANEUVERING Geos AND TIME ea LEAVING INITIAL RELATIVE: POSITION. Example.—At 0800 Guide is on course 140°, speed 14.0 knots, and will change to course 070°, speed 16.0 knots, at 0930. | Ship M, now stationed 8.0 miles on the port beam of the Guide, is ordered'to take station 12.0 miles on the starboard beam of the Guide, arriving at 1200, time of departure optional.’ M decides to maintain present Station as long as possible so as to carry out his orders at 14.0 knots. Required.—(a) Course of M.- (b) Time M leaves present Station. (c) Minimum speed and corresponding course avail- able to M.° (See fig. 29.) ; HLOOS : Ptocedure.—Lay out e..'/. g, ande... . g:, the first and second vectors of the Guide. Join g, and’ g). Plot the position of the Guide at any point G. “Eeeace Mj, the initial position of M,’ which is 8.0 miles bearing 050° from G. Inasimilar manner locate M, bearing 160° and distant 12.0 miles from G. | From the rules listed‘on the preceding page, the vector of the Fictitious Guide coincides with the vector of G’s first Test and the final position of M must be offset for the time that the real Guide and the Fictitious Guide are not together. Offset JM, in the direction parallel to sf . . $2, a distance equal to the Relative Run for 150 minutes at Relative Speed st. + §, as found from the Logarithmic Scale, locating M,’. Join M, and M,’, and transfer the slope of M,'.. .’. M2" to af loge the 14.0 knot circle at m. ‘The course for M at 14.0 knots is indicated by e'. . . .' m. | Using the Relative Speed sf. .'. . m and the Relative Distance M, .... M2’, the time required for M to esanghited the maneuver at 14.0 knots is fetbhd! This time, subtracted from 1200, gives file time at which M should first head for his final position. If the minimum speed is to be used, M must start at once. His course and speed would either be found by a simple Navigational Plot, or, using the offset position previously plotted, the Relative Speed would be the distance M, ...% . M,’ divided by the 4.0 hours available. This Relative Speed, gf . . . . m’, indicates that the minimum speed and the corre- sponding course is given by the vectore .... m’. Answer.—(a) 098°.’ ‘(b)’ 163 minutes before 1200 or at 0917. (c) 13.2 knots on course 111°. NOTE.—The procedure would be the same were the course and not the speed specified, the determining intersection being that of the slope from Sf parallel to M, . . . . M2’ with the specified course line. In case e .. .'. m! is not normal to gf . . | . m; it is possible to use a lower speed where these conditions obtain: However, the problem will have to start earlier than 0800 in order to utilize this lower speed. 56 Washington, D.C., published May, 1920, at Hydrographic Office, under the authority of the SECRETARY OF THE NAVY New Publication: 4th. Ed, Apr., 1938 4th. Edition, April 1938 No, 26659 50 ° = Sehles) i80 A tog te SCALES 2:1 3:1 : Jesuits Miles 4:1 5:1 38 57 s\aphon I Division =2 Knots 95 36—| 54 90 34 St 85 32— 48 80 30— 45 7S 28—| 42 70 26— 39 65 5 “st. 24— 36 5 a @ RELATIVE PLOT: pailles fees 3) 20-30 50 1— 27 45 16— 24 40 4 2i 5 i2— 18 30 wb 15 25 8— 12 20 6-9 15 4 6 10 ral &! 5 o— 0 0 328 8 g DISTANCE in yards Gi two corresponding quantities, solve Relative or actual iven any os in mit for third by laying rule through points on prop- & DISTANCE in miies er scales and read intersection on third scale. MANEUVERING BOARD SPEED in knots Relative, or actual Price 60 cents (per pad of 50) Weshington, D.C., published May, 1920, at Hydrographic Office, under the authority of the SECRETARY OF THE NAVY New Publication; 4th, Ed. Apr, 1938 4th. Edition, April 1938 No. 2665 a FIGURE 29. . 57 593046°—44——_5 Case XXi TO PROCEED FROM ONE RELATIVE POSITION TO ANOTHER AT KNOWN SPEED, STARTING AT KNOWN TIME, GUIDE CHANGING COURSE AND SPEED DURING OPERATION GIVEN: COURSES AND SPEEDS OF GUIDE AND TIME ON FIRST LEG, INITIAL AND FINAL RELATIVE POSITIONS OF MANEUVERING UNIT, SPEED OF MANUEVERING UNIT, AND TIME MANEUVER STARTS. TO DETERMINE: COURSE OF MANEUVERING UNIT AND TIME TO ARRIVE IN FINAL POSITION. Example.—Guide is on course 320°, speed 15.0 knots, and has signaled that at 1400 her course and speed will be changed to 280° and 12.0 knots. Ship M, now located 18.0 miles broad on the port beam of the Guide, receives orders at 1130 to take position 10.0 miles bearing 055° from the Guide as soon as possible using 15.0 knots. Required.—(a) Course for M. (b) Time required to reach final position. (See fig. 30.) Procedure.—Plot the Guide at any point G, and locate the initial and final positions of M at M, and M),, respectively. Measure M,.... Mp. Drawe....gande.... g, the vectors of G’s first and second legs. A rapid inspection of the plotted positions will indicate that it will be impossible for M to reach its new station in the 2.5 hours that G is on the first leg, since M is held to the same speed as G and must shift from 18.0 miles on one beam to 10.0 miles slightly abaft the other beam. This may be checked by transferring the slope M, .... Mz to gi, cutting the 15.0 knot circle at g; and m’. WRelative Speed g, .. . . m’ is too low to permit the change of station while G is on her first leg, so a Fictitious Guide must be used. Following the rules formulated in the introductory pages of this section, the Fictitious Guide takes the second vector of G, so that e..... sf coincides with e .... gs, and the initial position of M is offset for the time that the real Guide and the Fictitious Guide are not together. By means of the Logarithmic Scale, using the time G is on the first leg, and the Relative Speed g, ... . Sf, the distance M, is to be offset is obtained, thus locating M,’. Connect M,’... . M2 and transfer this slope to g2, cutting the 15.0 knot speed circle at m. e.... mis the course for M. By means of the Logarithmic Scale, the time required for the maneuver is found. Answer.—(a) 347°. (b) 195 minutes or 3.25 hours. NOTE.—Had minimum speed for M been a requirement instead of minimum time at 15.0 knots, the vector for M would have been found by dropping a perpendicular from e to the slope 8: .... m. Had the speed for M been between ee . . . . $2 and the minimum speed as explained above, and M was not required to be in position as soon as possible, two solutions would result through the transferred slope My’... . M2 crossing the specified speed circle twice. A Navigational Plot, to reduced scale, is inserted on the diagram to give a graphic picture of the tracks of the real Guide, the Fictitious Guide, and the Maneuvering Unit. 58 SCAR Pp ‘she ; a “= "T pivision=4°Miles . , i : ante | Division =2 Knéts 2:1 3:1 “NAVIGATIONAL PLOT ~ (Reduisza Scato) *, DISTANCE in yerds ¢ T Relative or actual Given any two corresponding quantities, solve v 2 R 8 DISTANCE in miles: for third by laying rule through points on prop- ; er scales and read intersection on third scale. MANEUVERING BOARD wlan itil Pa (een rein SPEED in nos Price 60'conts (por pad of 50) Now Publication: 4th. Ed, Apr, 1938 th, Edition, No. 2665 a. FIGURE’ 30: SECTION III TWO COURSES FOR MANEUVERING UNIT, SINGLE COURSE FOR GUIDE (THE FICTITIOUS SHIP) Tn section IL we dealt with a Fictitious Guide which, while maintaining a single course and speed, accompanied the actual Guide for an indefinite time while the latter was on either its first or its second course. One feature of this Fictitious Guide is that it maintains constant bearing with the real Guide, even when the two are not together. This is true because the two units either met or else departed from a common point. The Fictitious Guide was necessary in the solution of the problems listed in section II because we either did not know when the problem started or else we did not know when it terminated. In the cases to be investigated in this section, we do know when the problem starts as well as when it terminates. We know, also, the course and speed of the Guide and the initial and final relative positions of the Maneuvering Unit. The Maneuvering Unit, however, is not going directly from the initial position to the final position, but it is going to some inter- mediate position enroute. We therefore have two Lines of Relative Movement to consider instead of the one usually involved. In some cases these Relative Movement Lines may be specified chart lines. Although we know the total time of the problem, we do not know when the Maneuvering Unit must reach the intermediate point until we have partially solved the problem. To facilitate the solution of problems of this type, another fictitious unit is introduced. This unit is called the Ficititious Ship and will be used in both surface and aerial problems. The Fictitious Ship leaves the initial relative position simultane- ously with the Maneuvering Unit, proceeding directly for the final relative position at such a speed that the time of arrival of the Ficititious Ship coincides with the time of arrival of the Maneuvering Unit. The latter, meanwhile, has passed through the required intermediate point. Since the Fictitious Ship and the Maneuvering Unit leave the initial point at the same instant, they maintain constant bearing while the Maneuvering Unit is proceeding to the intermediate point. Also, since these two units come together again at the final position, they maintain constant bearing while the Maneuvering Unit is proceeding from the intermediate point to the final position. The bearing of the Fictitious Ship from the Maneuvering Unit (and the reverse), therefore does not change during the problem. It is this feature which gives the Fictitious Ship its particular value in the solution of two-course problems. Another feature, intimately connected with the Fictitious Ship, is called the Time Line and is illustrated in figure 31, which shows the Vector Diagram, the Relative Plot, and the Navigational Plot. In this sketch, both the Fictitious Ship and the Maneuvering Unit are originally at the point A. The Maneuvering Unit steers course 050° for 2.0 hours at a speed of 12.0 knots until point P is reached, when course is changed to 320° and speed to 18.0 knots. This latter course and speed are continued for 1.0 hour, at the end of which time the Maneuvering Unit has arrived at point B. The Fictitious Ship, which will be designated as F leaves point A and heads directly for point B, using course 013° and speed 10.0 knots. It will be noted that F leaves point A and arrives at point B simultaneously with the Maneuvering Unit M. As M proceeds from point A to point P in 2.0 hours, F runs from A to point F;; therefore, at the end of 2.0 hours, the bearing of Ffrom MisP ... . F,, and this bearing has not changed since the two units left point A. When M turns at point P, and heads for point B, F continues along the line F, . . . . B. The bearing between M and F does not change, therefore, and the two units would be in collision at point B. Using the Vector Diagram, where the vectors for the first and second legs run by Maree ....mande... .merespectively, theslopeP . . . . F, when transferred to m, is found to also pass through m, and f, the extremity of the vector for F. This, of course, is as it should be, because the bearing between F and M does not change during the entire problem. At the end of 2.0 hours, M has departed a distance F, .. . . PfromF. This distance divided by the time involved yields a rate which, when laid off on m, . . . . ™:, is found to coincide with m, . . . . f. Since M must again coincide with F at the end of the problem, or in 1 hour, the relative rate of approach between M and F must be P .. . . F, divided by 1.0 hour. Laying off this rate from m2, we see that it indicates the vector m, ... . f. From this, it is apparent that the line connecting the two vectors of M also includes the end of F’s vector, and that the rates m, ....fandm,... . fare directly proportional to the rates of departure and approach of the two units and are inversely proportional, to the times of departure and approach. The two units are departing from one another while M ison coursee . . . . m; and approaching while M ison coursee . . . . Ma, hence the line m, . . . . m» is divided by f into segments which are inversely proportional to the times M spends on the first and the second courses. m,....fand m.... fare relative vectors because the two units are on the courses indicated for M and for F simultaneously. m, ... . m:isnotarelative vector, but is what is described as a Time Line, so called because of the above-mentioned time division determined by f. For convenience, the characteristics of the Fictitious Ship may be summarized as follows: 1. It proceeds at constant speed and on a single course, from the initial position to the final position, leaving the former and arriving at the latter coincidentally with its reference ship, or unit, which reference ship or unit meanwhile passes through some intermediate point. 2. It maintains constant bearing with its reference unit throughout the operation. 3. The Fictitious Vector divides the line connecting the two vectors of the Reference or Maneuvering Unit into segments. inversely proportional to the times spent by the Maneuvering Unit on its two courses. 4. The Fictitious Ship can be used only for the solution of problems in which both time of start and time of finish are known. The above characteristics of the Fictitious Ship will become familiar during the solution of the two-course problems pre- sented in illustrating the cases for this section. A number of scouting problems are included, involving both surface craft and aircraft. During these problems when minimum speed is required, the earlier the Maneuvering Unit leaves its initial position the lower will be the speed necessary. Also, to use minimum speed, the same speed must be used on both legs. 60 Re ae pepe Ny ‘I Division =3' Miles’ 38. 57 p ao, eee: Mi oes nice Seer a Ne : £ 7+, 1 Division= 3 Knots Seep eeeed “RELATIVE PLOT 20—30 : ¥ : ie ee ieee 4 =a vd : : cong sob bae goed : : ; ; ; ae YEcTOR DIAG! Setar a 4 2i1- is “NAVIGATIONAL PLOT: sf Jaton. BB | 4 6 ee he = Wiehe’ ? pis Tpooeecsaceby ft =A ork - *: oa eet ss! 8 n = 4 ” z= 8 82 ame “ ” - » © RnR @oOae 2 Oe GREE CE WO Va eal RN RE g TIME in minutes 2 a 9 #8 8 sgaegscesseg 3 Segseegeseee 2 2 8% Given any two corresponding quantities, solve = “ a + 6 erero © ZSRE for third by laying rule through points on prop- er scales and read intersection on third scale. MANEUVERING BOARD i HNN Rarer Ma COMME ATPAAT ANT TTC PERCE REL enia ity tet teh y SEED in tts, Price 60 cents (per pad of 50) Washington, D.C., published May, 1920, at Hydrographic Office, undor the authority of the SECRETARY OF THE NAVY New Publication: 4th. Ed, Apr, 1938 4th. Edition, April 1938 No. 2665 a FIGURE 31. 61 Case XXII TO.PROCEED FROM ONE RELATIVE POSITION TO ANOTHER IN GIVEN TIME AT MINIMUM SPEED, PASSING THROUGH AN INTERMEDIATE RELATIVE POINT EN ROUTE GIVEN: COURSE AND SPEED OF GUIDE, INITIAL, INTERMEDIATE, AND FINAL RELATIVE POSITIONS OF MANEUVERING UNIT, AND TOTAL TIME INTERVAL. TO DETERMINE: COURSES AND SPEEDS OF MANEUVERING UNIT. Example.—Guide on course 040°, speed 12.0 knots, has ship M now stationed 8.0 miles broad on her starboard beam. M receives orders to take station 10.0 miles bearing 060° from the Guide, to pass through a relative point 25.0 miles bearing 085° from the Guide while shifting stations, and to complete the maneuver in 5.0 hours, using minimum speed. Required.—(a) Courses for M. (b) Speed of M. (c) Required time to reach intermediate position. (See fig. 32.) Procedure.—Plot the Guide at any convenient point G, and locate the initial, the intermediate, and the final positions of the Maneuvering Unit at M,, M2, and M3. Join M, and M2, M, and M3, and M, and M3. Lay offe.... g, the vector of the Guide.. Transfer the slope M, . .. . M, to g and mark it slope (1). Transfer the slope M, ... . Mz to g and mark it slope (2). The distance M, .... Ms; will be travelled by the Fictitious Ship in the 5.0 hours allotted. Therefore the Relative Speed of F is equal to M, . ... . M; divided by 5, and is plotted as g .... f, parallel to the slope of M, .... M3. The course and speed of F are te esented by. vector e).7.. 9h. Pivot a straight edge at f, and so orient it that it cuts slope (1) and slope (2) at m, and m, respectively, both of these points of intersection being equidistant from e. e ... . my is the first course for Mande... . mz is the course for the second leg. Speed is indicated by eithere ....m,ore.... mp». To obtain the time for J to reach the intermediate position, the Relative Distance M, ... . Mz, is divided by the Rela- tive Speed § . ... m,. Another way to obtain this time is by use of the Time Line m, ....f.... mp». Time on first Pages, tops Min yZI 29. leg is equal to — + times 5.0 hours. Mm, .. +. «© #No Answer.—(a) First course 048%°; second course 318°. (b) Speed 16.6 knots. (c) 4.1 hours. NOTE.—Since the minimum speed was specified, speed must be the same on both legs. The course of speed for M on either leg could have been specified, in which case the corresponding speed or course would have been found by the intersection of the specified course line or speed circle with the slope for that leg. A rapid method for finding the time on the first leg is to draw any line mz . . . . A from mz: equal in length to the total time on both courses, as measured on any convenient scale. Aand mareconnectedand/f.... Bisdrawnfrom/fparalleltom,;....A. mz... . Bthen measures to the same scale previously chosen the length of time on the first leg. 62 a 46k)66e!)~CUl OR Be Bseseeeees 48 8 geseesessss & 8 Given any two corresponding quantities, solve rs Se ge weer me Nae for third by laying rule through points on prop- er scales and read intersection on third scale. MANEUVERING BOARD 1 a mer oe tia SP TATE aac elle ia Price 60 conts (per pad of 50) Washington, D.C., published May, 1920, at Hydrographic Office, under the authority of the SECRETARY OF THE NAVY New Publication: 4th. Ed, Apr., 1938 FIGURE 32. | Division=2 Miles | Division= 3 Knots ggg 8 Bes s 8 @38 2 8 & 83 4th, Edition, April 1938 4:1 S:1 76-95 725-90 68 F-85 64 F-80 60 ;—-75 56-70 52 F-65 48 | 60 445-55 40-50 36 F—-45 32 F-40 28 F-35 24-30 20 F-25 16 20 12 15 8 E10 4.5 it) 0 g & TIME in minutes re tt DISTANCE in miles SPEED in knots Relative or actual No. 2665 4 63 Case XXIII TO SCOUT A GIVEN RELATIVE LINE AS FAR AS POSSIBLE FROM A GIVEN INITIAL RELATIVE POSITION, RETURNING TO ANOTHER RELATIVE POSITION IN SPECIFIED TIME AT SPECIFIED SPEED GIVEN: COURSE AND SPEED OF GUIDE, INITIAL AND FINAL RELATIVE POSITIONS, DIRECTION OF RELATIVE LINE, SPEED OF SCOUT, AND TOTAL TIME INTERVAL. TO DETERMINE: COURSES OF SCOUT, LENGTH OF RELATIVE LINE COVERED, AND TIME TO TURN TO FINAL COURSE. Example.—Fleet is on course 228°, speed 15.0 knots. Lexington, now located 85.0 miles due west from the Guide, re- ceives orders to launch a plane at 1200 for purposes of scouting a relative line 180° from the Lexington as far as possible, using air speeds of 90.0 knots and landing on the Saratoga at 1700. Saratoga is stationed 4.0 miles 42° abaft the starboard beam of the Guide. Wind is from 120°, force 20.0 knots. Required.—(a) Air courses for the plane. (b) Time plane changes course to head for Saratoga. (c) Length of relative line covered by plane. (d) Length of chart line covered on first leg. (See fig. 33.) Procedure.—Lay out e.... g, the course and speed of the Guide, which is also the course and speed of Saratoga. Lay out wind’s vectore .... w. Plot Fleet Guide at any convenient point, G, and locate relative positions of Lexington and Saratoga at L and S respec- tively. Lay out first Relative Line to be run by plane in direction 180° from L of indefinite length and transfer this slope to g. Join L and S, the distance run by the Fictitious Ship. Transfer this slope to § and lay out g .. . . f equal to the rate found by dividing L .... S by the 5.0 hours available. With w as center and radius equal to 90.0 knots, draw a circle of the plane’s air speed, intersecting the first slope from g at pi). Ww... . p; is the first air course of the plane, which makes good a course indicated on the chart bye... . py. Draw a time line from p, through f to the plane’s speed circle, intersecting at p>. w ... . Pp: is the second air course for the plane. Join p..... g and transfer slope to S, intersecting outgoing relative line from ZL at P, which is the Relative Position reached by the plane when course is changed to head for Saratoga. Time plane is on first leg is found by dividing Relative Distance L . . . P by ISSELEN Speeds ... pi, which time may also be found from the Time Line p, ....f... . p. as previously explained: Length of relative line covered is L . . P. Length of chart line covered is Ground Speed, e . . . . p;, multiplied by the time on first leg. Answer.—(a) First course 176!/,°; second course 0311/,°. (b) 1451. (c) 198 miles. (d) 230 miles. NOTE.—Solution for surface ships is the same except that wind vector and air speed circle are not used. The required speed circles are drawn about e, which is the origin of the courses to be steered. The speed on both legs need not be the same, but the Time Line must pass through f. The action of the wind results in the necessity of steering a course to the left of the ground course while on the first leg and the steering of a course to the right of the ground course on the second leg. (64 ae sees) - ae I Division =10 Miles acer 38— 57 oe | Division = 10 Knots 76-95 ay 36—| 54 fo 72/90 Jicres Jy: 4 5I at 68 F-85 | gree 32 48 I. 64-80 pte nil I 2a—| 42 4 56-70 26—4 39 52 F-65 24 36 | 48-60 g. | 233 | 44f-55 | | 20-30 tt S [40550 | 18— 27 36 F-45 6 24 32 F- 40 4 21 28 F-35 re 24-30 ob 15 20-25 8— 2 16 F-20 6 9 12 5 446 8 F-0 2 3 4 5 0 t) 0 it} e : z 3x8 8 g & 7 TIME in minutes 8 8 88gessh 2 8 g 8 gg gagghesses § 8 898888 88eee Fo eS EEg8 8 B pistANce inyans Relative or actual: = % CO AAC 2 8 RKRVIS ® $8 DISTANCE in miles = OT ” < » © ROO 2 g x &8 & g 3 SPEED in knots Price 60 cents (per pad of 50) Weshington, D.C, published May, 1920, at Hydrographic Office, ‘under the authority of the SECRETARY OF THE NAVY Now Publication: 4th. Ed, Apr. 1938 4th. Edition, April 1938 No. 2665 a FIGURE 33. 65 Case XXIV TO SCOUT A GIVEN CHART LINE AS FAR AS POSSIBLE FROM A GIVEN INITIAL POSITION, RETURNING TO ANOTHER RELATIVE POINT IN GIVEN TIME AT GIVEN SPEED GIVEN: COURSE AND SPEED OF GUIDE, INITIAL AND FINAL RELATIVE POSITIONS, DIRECTION OF CHART LINE TO BE SCOUTED, SPEED OR SPEEDS OF SCOUT, AND TOTAL TIME OF OPERATION. TO DETERMINE: COURSES OF SCOUT, LENGTH OF CHART LINE SCOUTED, AND TIME TO TURN. Example.—Carrier C, now on course 500° at a speed of 20.0 knots, is 60.0 miles, 110° from an air station. A plane takes off from the air station at 0600 to scout a chart line in direction 140° as far as possible, landing on the carrier at 1000, using an air speed of 85.0 knots for the entire period. The true wind is from 340°, velocity 25.0 knots. Required.—(a) Air courses for the plane. (b) Length of chart line scouted. (c) Time plane turns to head for carrier. (d) Bearing and distance of carrier from plane at (c). (See fig. 34.) Procedure.—Lay oute ....cande... . w, vectors of carrier and wind, respectively. With w as center, draw plane’s air speed circle, radius 85.0 knots. Plot the air station at any convenient point, AS, and locate the present relative position of the carrier at C. Lay off first chart course of plane from e in the specified direction, 140°, intersecting the plane’s speed circle at p;. Join w....piandc.... p;. The first air course for the plane isw....p:.. The vectorc.... pi indicates the’ slope of the Relative Movement Line of the plane on its first leg. Join AS .... C and transfer slope to c. Along this slope lay off rate AS ... . C divided by total time 4.0 hours, locating point #. The Relative Speed and course of the Fictitious Ship is indicated by this vectorc.... f. Draw Time Line p, ....f... . Do, intersecting the plane’s speed circle at p.. Ww... . p2 is the second air course of plane. Transfer the slopec .... p, to AS and the slope cc... . p, to C. These slopes intersect at P, the turning point for the plane. The time on the first leg is found by dividing the Relative Distance AS by the Relative Speed c .... p; or else by using the Time Line and the formula. Time on first leg equals f .. ~~. De divided by p:....f.... p2times Total Time Allotted. The length of chart line covered is found by multiplying the Ground Speed, e . . . . p;, by time on first leg. Answer.—(a) First course 1341/,°; second course 358°. (b) 162 miles. (c) 0729. (d) 349° distant 123 miles. NOTE.—To illustrate the scouting of a line in a given chart direction by a surface ship, example XXIV-B is appended. 66 "[ Division =10 Knots | Division=10 Miles TIME in minutes DISTANCE in yards Relative or actual Given any two corresponding quantities, solve “2 2£ & 8ggsezs & -§ DISTANCE in miles for third by laying rule through points on prop- er scales and read intersection on third scale. SPEED in knots NUMRIRUIRIRTNGHRO RID, ct Se Tee eM i ate eae een A alae S22 ios, Price 60 cents (per pad of 50) Washington, D.C., published May, 1920, et Hydrographic Office, under the authority of the SECRETARY OF THE NAVY Rew Publication: 4th, Ed., Apr, 1938 4th. Edition, Apri 1938 No, 2665 a FIGURE 34. 67 Example XXIV-B GIVEN: Fleet Guide is on course 340°, speed 16.0 knots, and at 0800 is 20.0 miles bearing 250° from chart point A. At this time a destroyer leaves point A to scout a chart line in direction 000° to maximum distance so as to join the Guide at 1300, using a speed of 25.0 knots on the first leg and a speed of 20.0 knots on the second leg. Required.—(a) Destroyer’s course on first and second legs. (b) Length of chart line scouted. (c) Time Destroyer turns to second leg. (See Fig. 35.) Procedure.—Locate the chart point at any point, A, and the 0800 position of the Guide at G. Lay out the Guide’s vector,e. .. . g, and the vector of the first leg for the Destroyer,e.... d,. Lay off from g the vector... . f, equal in length to the rate obtained by dividing the distance A. . . . G by the 5.0 hours available, and parallel to the slope A... . G. From d, draw a Time Line through f, intersecting the 20.0 knot speed circle at d2. e@. . . . ch is the vector for the second leg of the destroyer’s run. From A, in direction parallel tog... . di, lay offaline A... . X, of extended length. From G, lay off a line parallel to, but in opposite direction to, 8. . . . d:, intersecting A.... XatP. A....PandP.... G represent the Relative Lines run by the destroyer on its first and second legs respectively. The time on first leg may be found either by dividing the Relative Distance A. . . . P by the Relative Speed §.... dh, or by proper use of the Time Diagram. Time to change to second leg is this time added to 0800. The chart distance run on course 000° is found by multiplying the speed of 25.0 knots by the time on the first leg. Answer.—(a) First course 000°; second course 272°. (b) 67.0 miles. (c) 1041. NOTE.—A Navigational Plot of this example is shown in half scale for illustrative purposes, but is not required for the solution. A comparison of this example of surface vessels with the previous example using airplanes, will further emphasize the superiority of aircraft for covering large areas in scouting exercises. 68 “| Division =3 Knots | Division=5 Miles : male : J Bn F 76-95 Reale f Pe | ee Bae 5s BAe omy 7290 to; eens 2 We é Reed ee :. 68 F285 Me P ‘ re a ; Mi a J5 o ery coe Lees Nes lipase ASE aC alc 56 F-70 Soot baat pao SERS, HO AES 2 - phe : 52-65 pl Kola They shy ite sce sae RN TY F i 4455 - NAVIGATIONAL-PLOT 7"... g : : BE : IDivision=16 Miles. if iy fate : OE ee ay ee 3 fe r a5 we oteg, - B35 i ao s ‘s 4 as y i oe “s., ao fe 7 i 20-25 ee ‘ cal he zat : 16 F-20 © 8g DISTANCE in yards Relative or actual DISTANCE in mies 2 2 gggegghesses 8 8 888 Given any two corresponding quantities, solve . i ore for third by laying rule through points on prop- er scales and read intersection on third scale. MANEUVERING BOARD 1 i Woshington, D.C., published May, 1920, at Hydrographic Office, Price 60 cents (per pad of 50) under the authority of the SECRETARY OF THE NAVY New Publication: 4th. Ed, Apr, 1933 4th. Edition, Apcit 1938 No. 2685 0) FIGURE 35. 15 0. n ” < » © R@D@aES SPEED in knots Relative, or actual 69 Case XXV TO SCOUT A GIVEN RELATIVE LINE AS FAR AS POSSIBLE SO AS TO REACH A GIVEN CHART POINT IN GIVEN TIME AT SPECIFIED SPEED GIVEN: COURSE AND SPEED OF THE GUIDE, INITIAL RELATIVE AND FINAL CHART POSITIONS, DIREC- TION OF RELATIVE LINE TO BE SCOUTED, SPEED OF SCOUT, AND TOTAL TIME INTERVAL. TO DETERMINE: COURSES OF SCOUT, LENGTH OF RELATIVE LINE COVERED, LOCATION OF TURNING POINT, AND TIME OF TURNING TO SECOND COURSE. Example.—Lexington, on course 315° at speed 25.0 knots, at 0615 bears 210°, distant 70.0 miles from a Naval Air Station. At 0615 a plane is launcHed to scout a relative line in direction 340° from the carrier, as far as possible, returning to the Naval Air Station in 4.0 hours. Plane has air speed 70.0 knots; and the wind is from 100° with a velocity of 22.0 knots. Required.—(a) Air courses for the plane. (b) Relative length of line covered. (c) Bearing and distance of Lexington when plane starts second leg. (d) Time plane starts second leg. (See fig. 36.) Procedure.—Lay out vectors of Lexington and wind,e ....ilande.... w,respectively, and draw 70.0 knot speed circle with w as center. Plot position of Lexington at any point, L, and locate Naval Air Station at N, 70.0 miles bearing 030° from Lexington’s 0615 position. From J, lay out the slope of the required Relative Movement in direction 340°, intersecting the plane’s air speed circle at pi. The first air course for the planeisw .... Di. Draw EL... . N,and transfer slope to e, since L . . . . Nrepresents an instantaneous chart position at 0615. Lay out along this slope the rate obtained by dividing distance L . . . . N by total time, 4.0 hours, locating f. e... . fis the vector of the Fictitious Ship. From p,; draw a Time Line through /, intersecting the plane’s speed circle at p2. W .. . . P2is the course for the second leg. Using the 0615 position of Lexington, the charted position NV, and the courses made good over the ground by the plane, e....prande.... ps, the chart track of the plane is from L to P and back to NV. Maintaining L as the fixed position of the Guide, the position of NW must be offset an amount equal to the Relative Move- ment of N during the 4.0 hours allowed. This is in direction 7... . e and locates N’, the position of the Naval Air Station relative to Lexington at 1015. Transfer theslopeof /....p,toLandthe slope of 1... .p,to N’. These two slopes intersect at T, the turning point for the plane. L.... Tis the relative length of line covered before changing course. 7....L is the bearing and distance of Lexington from plane at turning point. Time on first leg is found by dividing L...).. Tbyrate]....p:,0.... Pby ground speede.... p,, or by use of the proportions of the Time Line p, ....f.... p2. This time added to 0615 gives the time to turn. Answer.—(a) First course 347°; second course 114°. (b) 95.5 miles. (c) Bearing 160°, distant 95.5 miles. (d) 0755. NOTE.—The solution of this example would be the same for a surface scout except that the courses would be referred to e instead of w. Also speeds need not be the same on each leg. 70 _ I Divisjon=10 Mites - *., 1 Division=10 Knots _ DISTANCE in yards _ Relative or actual 2 8 & ARSSss DISTANCE. in miles Given any two corresponding quantities, solve for third by laying rule through points on prop- er scales and read intersection on third scale. : = “ ” + » © » @oe » .&8 8 8 8 8B § SPEED in knots “MANEUVERING BOARD 5 | L Sea are re rr I TET Fee Relative or actual Price 60 cents (per pad of 50) Weshington, D.C., published May, 1920, at Hydrographic Office, under the authority of the SECRETARY OF THE NAV New Publication: 4th. Ed., Apr 1938 4th. Edition, April 1933 No. 2685 @ FIGURE 36. a1 Case XXVI TO PROCEED FROM ONE RELATIVE POSITION TO ANOTHER IN GIVEN TIME AT MINIMUM SPEED, PASSING THROUGH A GIVEN CHART POINT EN ROUTE GIVEN: COURSE AND SPEED OF GUIDE, INITIAL AND FINAL RELATIVE POSITIONS, RELATIVE LOCA- TION OF CHART POINT, AND TOTAL TIME OF OPERATION. TO DETERMINE: COURSES AND SPEEDS OF MANEUVERING UNIT, TIME OF REACHING CHART POINT, AND RELATIVE POSITION OF TURNING POINT. Example.—Fleet is on course 140°, speed 12.0 knots. At 0630 a plane is launched from the Saratoga with orders to investigate a chart point bearing 200° and distant 105.0 miles from Saratoga and to land on the Ranger, which is located 45.0 miles ahead of Saratoga, and is on fleet course at fleet speed. The plane is to complete the maneuver at 0930, using mini- mum speed en route. Wind is from 010°, velocity 21.0 knots. Required.—(a) Minimum air speed for plane. (b) Air courses for plane. (c) Time chart point is reached. (d) Bearing and distance of plane from Saratoga when chart point is reached. (See fig. 37 ») Procedure.—Lay out the wind’s and the Saratoga’s vectorsase.... W ande. ; z= Ss, respectively. The latter is also the vector of the Ranger, which is also using Fleet Course and speed. At any convenient point lay out the position of the Saratoga at S, and the relative position of the chart point at 0630 at P. Locate the initial position of the Ranger at R. Advance the position of R to R’, representing 0930 position of this ship, 36.0 miles ahead of the 0630 position. Connect S and P and transfer this slope toe. Ina similar manner, connect P and R’ and transfer this slope toe. The plane travels the chart track S... - IPP RE ERS , Divide the distance S. . . . R by the 3.0 hours allowed, giving the rate travelled by the Fictitious Ship in its direct — route from the Saratoga to the Ranger. Froms,draws.. .- - f, parallel to S . . . Rand equal to the rate obtained — above. | Pivot a straight-edge at f and orient it until it cuts the slopes drawn from e and parallel to Sy uy ePyand! Pe . senha respectively, at points equidistant from w, these points being p; and pz, respectively. The air speed for the plane is given by | either w.... ..PyOrw. .:... px» The firstcourse for the plane is indicated by the direction of w. . . . pi and the) second air course byw. .. - P2- | From S, draw a Relative Movement Line parallel to the slope ofs. . . - Pi and from R, draw a Relative Movement Line parallel to the slope of p2. - . - ©. These two lines intersect at P”, the turning point. The time required to reach this point is either found by dividing the Relative Distance SS... . P” by the Relative Speeds... -. pi Or by the use of the Time Line p, . . .- f.. . . Ds, as previously explained. The bearing and distance of P” from S is the bearing and distance of the plane from the Saratoga at the turning point. Answer.—(a) 66.8 knots. (b) First course 203°; second course 05114°. (c) 0740. (d) Bearing 207°, distant 96.0 miles. NOTE.—Except for the influence of the wind, this case is most rapidly solved by the Navigational Plot, the method used for surface vessels. It will be noted that the Navigational plot is shown THES o.g'0 0 We oo R', while the Relative Plot is shown iy Oc o aeleeeouono ae 72 . ae | Division =10 Miles “+, 1 Division=10 Knots TIME in minutes gggses 8 88esss8 8 8 ne © &® & @ aeeese Sees 22 SeSetaeeee Fs ceece & & nemace myer Given any two corresponding quantities, solve : 3 me G3 DO tha) } 2 anand 33388 2 8 CRoraNGe ries for third by laying rule through points on prop- a : er scales and read intersection on third scale. MANEUVE - n ” z 7) o ne @ ae 2 8 8 8 8 ¢ 3 3 SPEED in knots RING BOARD [EE ee nt inn thivitin Relative or actual Price 60 conts (per pad of 50) Weshington, 0.C., published May, 1920, at Hydrographic Office, under the authority of the SECRETARY OF THE NAVY Now Publication: 4th. Ed. Apr. 1933, 4th. Edition, April 1933 No. 2665 a FIGURE 37. 593046°—4i——6 73 Case XXVII TO SCOUT OUT AND IN ALONG GIVEN RELATIVE LINE AT MINIMUM SPEED IN GIVEN TIME, RETURNING TO ORIGINAL RELATIVE POSITION GIVEN: COURSE AND SPEED OF GUIDE, INITIAL RELATIVE POSITION (WHICH IS ALSO FINAL RELATIVE POSITION), DIRECTION AND LENGTH OF RELATIVE LINE, AND TOTAL TIME INTERVAL. TO DETERMINE: COURSES AND SPEED OF SCOUT AND TIME TO TURN. Example—Guide on course 320°, speed 12.0 knots, has a scout stationed 4.0 miles broad on her starboard bow. Scout receives orders to scout, at minimum speed, to a distance of 25.0 miles from her present station, maintaining constant bearing on the Guide, leaving formation at 1300 and returning to station at 1600. Required.—(a) Speed used by Scout. (b) Courses for Scout. (c) Time to turn to second course. (See fig. 38.) Procedure.—Plot the Guide at point G, and locate the initial position and the intermediate point of the scout at S and P respectively. Join Sand P. S.... P is the Relative Movement Line on the first course and P ... . S is the Relative Movement Line for the second course. Lay out the Guide’s vector,e ... . g, and transfer the slope S . . . . P to g, extending in both directions. The use of the regular Fictitious Ship in this case would involve a vector g . . . . f of length equal to zero, since the initial and the final positions coincide, and yielding an indeterminate result. This is overcome by introducing a second Fictitious Ship whose vector is equal to the length of the perpendicular erected from e to the transferred slope S.... P. Relative to this extra Fictitious Ship the Fictitious Scout travels at the same rate both-going out and returning. Mathematically this constant rate is equal to the Relative Distance divided by the total time available plus the square root of the sum of the squares of the Relative Distance divided by the total time available and of the Relative Speed of the Guide to the Second Fictitious Ship. This process is laborious, so the procedure outlined below will be used as the graphic solution. Erect a perpendicular at e to the transferred slope, cutting it at O. Extend this perpendicular beyond the slope line. Along this same perpendicular lay out O . . . . @ equal to the Relative Distance divided by the total time available and to the speed scale in use. Connect @.... g and lay out length @ . . . . g from @ along the perpendicular, locating point R. With O as center and radius equal to O . . . . R, swing an arc cutting the transferred slope at s; and s,. The first course is e..... 8; and the second course is e . .. . S. The speed is shown by the length of eithere ....s,ore....& Time on first leg is found by dividing Relative Distance S . . . . P by Relative Speed §.... S. Answer.—(a) 22.0 knots. (b) First course 342°; second course 20712°. (c) 1508. NOTE.—The solution of this case using aircraft is shown in example XXVII-B. 74 in minutes 8 g DISTANCE in yards Relative or actual 3 DISTANCE in miles 75 Given any two corresponding quantities, solve for third by laying rule through points on prop- er scales and read intersection on third scale. SPEED in knots MANEUVERING BOARD Relative or actual Price 60 conts (per pad of 50) Washington, D.C., published May, 1920, at Hydrographic Office, under the authority of the SECRETARY OF THE NAVY New Publication: 4th, Ed, Apr, 1938 4th. Edition, April 1938 No. 2665 a , FIGURE 38. 75 Example XXVII-B Carrier C, on course 140°, speed 20.0 knots, launches a plane at 1115 to scout a relative point bearing 210°, distant 100.0 miles, returning in 2.5 hours. Wind is from 170°, velocity 25.0 knots. The maneuvers to the Carrier to launch the plane may be disregarded. Required.—(a) Air speed of plane. (b) Air courses for plane. (c) Time to turn to homing course. (See fig. 39.) Procedure.—Plot position of Carrier at any point C, and the relative point to be reached by the plane at P. Lay out the Carrier’s vector at e ....c and the wind’s vector ate ....w. Transfer the slope C....Ptoc, extending in both directions. Since the wind is the controlling factor with all aircraft problems, erect a perpendicular to the transferred slope of C.... P from w, intersecting the transferred slope at O. Extendw .... O indefinitely. From O, lay out O.... @ equal to the Relative Distance, 100.0 miles, divided by the 2.5 hours available, and to the speed scale in use. Join @and Cand lay out@.... Requalto@.... Cinlength. With O as center and radius equal toO... . R, swing an arc cutting the transferred slope at p, and pp, respectively. The air speed of the plane is given by the length of w....p,0rw.... ps. First air course is indicated by the direction of w .. . . p, and the second air course by the direction asw .... po. Time on first leg is found by dividing the Relative Distance C .... P by the Relative Speedc ....p,. This time added to 1115 gives the time to turn. Answer.—(a) 95.0 knots. (b) First air course 188%°; second air course 051°. (c) 1252. NOTE.—It is immaterial whether the perpendicular extends toward e or away from e as the same points p; and pz will result in either case. The ground courses and speeds of the plane are indicated by vectorse ....piande.... p2. Ifit were required that a plane guard trail the plane, it would therefore take the courses indicated bye ....piande.... p». In case the transferred slope passes through e for surface vessels or w for aircraft, the perpendicular from e or w respectively would have zero length, but it would still have direction. The procedure outlined above is still carried out, considering that the point O would coincide with e or w. 76 __EDivision=i0 Miles. ~* «| Division=10 Knots ~. TIME in minutes gga 8 8 88988 8 § 2 3 g fase sus gaged 22888 8 8 8 § see8 $2882 8 @ 8828828 $$ & pistanck inyards diven a two correspondi tities, solve - Sena iNg qua! sO! = o ao fo ORD © ca] “ © ii fee ary ra Coreen peing entities te 2 & 883828 8 8 __ DISTANCE inmiles er scales and read intersection on third scale. - ss a SL Se ole ice > 8. 8 BBS 8 6 Sayers MANEUVERING BOARD L : iaitun! Relative or actual Price 60 cents (por pad of 50) Washington, D.C., published May, 1920, at Hydrographic Office, under the authority of the SECRETARY OF THE NAVY. Now PubitooBor: 4th. Ed, Apr. 1933 401 Edition, /oril 1938 No. 2665 a FIGURE 39. 77 Case XXVIII TO SCOUT IN A GIVEN RELATIVE DIRECTION FOR MAXIMUM DISTANCE AT GIVEN SPEED, RETURNING TO ORIGINAL RELATIVE POSITION IN GIVEN TIME, USING GIVEN SPEED (OR SPEEDS) GIVEN: COURSE AND SPEED OF GUIDE, COINCIDENT INITIAL AND FINAL RELATIVE POSITIONS OF SCOUT, DIRECTION OF RELATIVE LINE TO BE SCOUTED, SPEED OF SCOUT, AND TOTAL TIME INTERVAL. TO DETERMINE: COURSES OF SCOUT, TIME TO TURN TO SECOND COURSE, AND LENGTH OF RELA- TIVE LINE SCOUTED. Example.—A Carrier is on course 090°, speed 17.5 knots, and at 0600 launches a plane to scout a relative line in direction 150° from the Carrier, as far as possible at an air speed of 85.0 knots, returning to the Carrier at 0820. Wind is from 110°, velocity 18.0 knots. Required.—(a) Air courses for the plane. (b) Time to turn to incoming course. (c) Length of relative line covered. (See fig. 40.) Procedure.—Plot Carrier at any point, C, and lay out the Relative Line C .... X in direction 150° from C and of indefinite length. Lay oute ... . c, the vector of the Carrier, ande .... w, the vector of the wind. With w as center and the airspeed of the plane as radius, inscribe a circle. Transfer the slope of C. . . . X toc, cutting the plane’s airspeed circle at p; and p,. The plane’s outgoing air course is Ww... . p, and the incoming air course isw... . Dr». The time on the first course is either found by means of the Time Line p, ....c.... p or else by the graphical method explained in case XXI. This time is added to 0600 to find the time to turn. The length of Relative Line scouted is found by multiplying the Relative Speed, c . . . . pi, by the time on the first leg or the Relative Speed, c . . . . po, by the time on the second leg, and drawn asC.... P. Answer.—(a) First air course 1311/,°; second air course 348°. (b) 0730. (c) 87.0 miles. NOTE.—With both the direction of relative movement and the plane’s airspeed given, it is a simple matter to find the desired air courses. The utilization of the Time Line provides a ready solution for the time on first and second legs. As in case XXVII, the Fictitious Ship’s vector coincidws with the Guide’s vector since the initial and final relative positions of the scout are the same. Thelinep; ....c... . pz is therefore a Time Line. The Relative Plot is not required if the scout leaves and returns to the Guide, but it is necessary otherwise if it is desired to know the scout’s position relative to the Guide at the turning point. In case the scout is a surface vessel, the speed circles are drawn from e instead of w. 78 50. ie} = SCALES oy Ls mod Wine Ao eters wy. 8 Zales! a Re fo Paci : : $ ¢ °+._ -[Division=10 Miles, 38 57 ue LSP Van onset ; ‘ m\ eae PAE OP COR ne : ; *+ | Division=10 Knots’: TIME in minutes g 288 8 2 8328 % DISTANCE in yards t Relative or actual Given any two corresponding quantities, solve = “ > © ©» oRea9 © 8288328 2 8 __ DISTANCE inmiles for third by laying rule through points on prop- er scales and read intersection on third scale. - “ ™ . -b° Oo Re @OHe 2. & 8 8 38 ¢ 8 3 SPEED in knots Price 60 cents (por pad of 5D) Washington, D.C., published May, 1920, st Hydrographic Office, Sean } under tha authority of the SECRETARY OF THE NAVY Wow Publication: 4th. Ed. Apr. 1938 4th. Edition, April 1938 No, 2085 a FIGURE 40. O 79 ios Rieti tips hia: femora Bia oan 4 ie Siieb 0 D6 ie iP mf Linay sg ‘ive Bt i ae mee ; Mees lJ » ora. ©) eee) decors Pye toe s ather Ce ithie oat chen ) eM a Vg wines 3 the] ye rbie eke Selec atane ea ee een a he bthiedt sii sant me, jioenetcbyag tek tee. 7. hy’ cy BT ans 2 Ss ala sare eo wg 5 « ' 6 Sd) > eA i ‘4 ay) by A. “ bg ec iy ttaed itu Aisne ns LW eakontalh oH eet ew cee rs a ae Su Bete BH oh)